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CyC2018
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@ -15,14 +15,14 @@
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* [分治](#分治)
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* [动态规划](#动态规划)
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* [斐波那契数列](#斐波那契数列)
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* [矩阵路径](#矩阵路径)
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* [数组区间](#数组区间)
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* [分割整数](#分割整数)
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* [最长递增子序列](#最长递增子序列)
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* [最长公共子序列](#最长公共子序列)
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* [0-1 背包](#0-1-背包)
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* [数组区间](#数组区间)
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* [股票交易](#股票交易)
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* [字符串编辑](#字符串编辑)
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* [分割整数](#分割整数)
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* [矩阵路径](#矩阵路径)
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* [其它问题](#其它问题)
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* [数学](#数学)
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* [素数](#素数)
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* [最大公约数](#最大公约数)
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@ -2477,6 +2477,258 @@ private int rob(int[] nums, int first, int last) {
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dp[N] 即为所求。
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### 矩阵路径
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**矩阵的最小路径和**
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[64. Minimum Path Sum (Medium)](https://leetcode.com/problems/minimum-path-sum/description/)
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```html
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[[1,3,1],
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[1,5,1],
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[4,2,1]]
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Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.
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```
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题目描述：求从矩阵的左上角到右下角的最小路径和，每次只能向右和向下移动。
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```java
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public int minPathSum(int[][] grid) {
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if (grid.length == 0 || grid[0].length == 0) {
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return 0;
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}
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int m = grid.length, n = grid[0].length;
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int[] dp = new int[n];
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (j == 0) {
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dp[j] = dp[j]; // 只能从上侧走到该位置
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} else if (i == 0) {
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dp[j] = dp[j - 1]; // 只能从左侧走到该位置
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} else {
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dp[j] = Math.min(dp[j - 1], dp[j]);
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}
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dp[j] += grid[i][j];
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}
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}
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return dp[n - 1];
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}
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```
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**矩阵的总路径数**
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[62. Unique Paths (Medium)](https://leetcode.com/problems/unique-paths/description/)
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题目描述：统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。
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<div align="center"> <img src="../pics//7c98e1b6-c446-4cde-8513-5c11b9f52aea.jpg"/> </div><br>
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```java
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public int uniquePaths(int m, int n) {
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int[] dp = new int[n];
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Arrays.fill(dp, 1);
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for (int i = 1; i < m; i++) {
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for (int j = 1; j < n; j++) {
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dp[j] = dp[j] + dp[j - 1];
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}
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}
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return dp[n - 1];
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}
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```
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也可以直接用数学公式求解，这是一个组合问题。机器人总共移动的次数 S=m+n-2，向下移动的次数 D=m-1，那么问题可以看成从 S 从取出 D 个位置的组合数量，这个问题的解为 C(S, D)。
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```java
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public int uniquePaths(int m, int n) {
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int S = m + n - 2; // 总共的移动次数
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int D = m - 1; // 向下的移动次数
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long ret = 1;
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for (int i = 1; i <= D; i++) {
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ret = ret * (S - D + i) / i;
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}
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return (int) ret;
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}
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```
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### 数组区间
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**数组区间和**
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[303. Range Sum Query - Immutable (Easy)](https://leetcode.com/problems/range-sum-query-immutable/description/)
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```html
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Given nums = [-2, 0, 3, -5, 2, -1]
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sumRange(0, 2) -> 1
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sumRange(2, 5) -> -1
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sumRange(0, 5) -> -3
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```
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求区间 i \~ j 的和，可以转换为 sum[j] - sum[i-1]，其中 sum[i] 为 0 \~ i 的和。
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```java
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class NumArray {
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private int[] sums;
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public NumArray(int[] nums) {
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sums = new int[nums.length];
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for (int i = 0; i < nums.length; i++) {
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sums[i] = i == 0 ? nums[0] : sums[i - 1] + nums[i];
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}
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}
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public int sumRange(int i, int j) {
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return i == 0 ? sums[j] : sums[j] - sums[i - 1];
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}
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}
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```
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**子数组最大的和**
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[53. Maximum Subarray (Easy)](https://leetcode.com/problems/maximum-subarray/description/)
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```html
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For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
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the contiguous subarray [4,-1,2,1] has the largest sum = 6.
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```
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```java
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public int maxSubArray(int[] nums) {
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if (nums == null || nums.length == 0) {
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return 0;
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}
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int preSum = nums[0];
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int maxSum = preSum;
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for (int i = 1; i < nums.length; i++) {
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preSum = preSum > 0 ? preSum + nums[i] : nums[i];
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maxSum = Math.max(maxSum, preSum);
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}
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return maxSum;
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}
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```
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**数组中等差递增子区间的个数**
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[413. Arithmetic Slices (Medium)](https://leetcode.com/problems/arithmetic-slices/description/)
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```html
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A = [1, 2, 3, 4]
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return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.
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```
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dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
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如果 A[i] - A[i - 1] == A[i - 1] - A[i - 2]，表示 [A[i - 2], A[i - 1], A[i]] 是一个等差递增子区间。如果 [A[i - 3], A[i - 2], A[i - 1]] 是一个等差递增子区间，那么 [A[i - 3], A[i - 2], A[i - 1], A[i]] 也是。因此在这个条件下，dp[i] = dp[i-1] + 1。
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```java
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public int numberOfArithmeticSlices(int[] A) {
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if (A == null || A.length == 0) {
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return 0;
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}
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int n = A.length;
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int[] dp = new int[n];
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for (int i = 2; i < n; i++) {
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if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
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dp[i] = dp[i - 1] + 1;
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}
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}
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int total = 0;
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for (int cnt : dp) {
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total += cnt;
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}
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return total;
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}
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```
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### 分割整数
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**分割整数的最大乘积**
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[343. Integer Break (Medim)](https://leetcode.com/problems/integer-break/description/)
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题目描述：For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).
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```java
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public int integerBreak(int n) {
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int[] dp = new int[n + 1];
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dp[1] = 1;
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for (int i = 2; i <= n; i++) {
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for (int j = 1; j <= i - 1; j++) {
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dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j)));
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}
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}
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return dp[n];
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}
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```
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**按平方数来分割整数**
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[279. Perfect Squares(Medium)](https://leetcode.com/problems/perfect-squares/description/)
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题目描述：For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
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```java
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public int numSquares(int n) {
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List<Integer> squareList = generateSquareList(n);
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int[] dp = new int[n + 1];
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for (int i = 1; i <= n; i++) {
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int min = Integer.MAX_VALUE;
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for (int square : squareList) {
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if (square > i) {
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break;
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}
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min = Math.min(min, dp[i - square] + 1);
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}
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dp[i] = min;
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}
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return dp[n];
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}
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private List<Integer> generateSquareList(int n) {
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List<Integer> squareList = new ArrayList<>();
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int diff = 3;
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int square = 1;
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while (square <= n) {
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squareList.add(square);
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square += diff;
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diff += 2;
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}
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return squareList;
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}
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```
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**分割整数构成字母字符串**
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[91. Decode Ways (Medium)](https://leetcode.com/problems/decode-ways/description/)
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题目描述：Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).
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```java
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public int numDecodings(String s) {
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if (s == null || s.length() == 0) {
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return 0;
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}
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int n = s.length();
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int[] dp = new int[n + 1];
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dp[0] = 1;
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dp[1] = s.charAt(0) == '0' ? 0 : 1;
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for (int i = 2; i <= n; i++) {
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int one = Integer.valueOf(s.substring(i - 1, i));
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if (one != 0) {
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dp[i] += dp[i - 1];
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}
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if (s.charAt(i - 2) == '0') {
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continue;
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}
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int two = Integer.valueOf(s.substring(i - 2, i));
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if (two <= 26) {
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dp[i] += dp[i - 2];
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}
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}
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return dp[n];
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}
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```
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### 最长递增子序列
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已知一个序列 {S<sub>1</sub>, S<sub>2</sub>,...,S<sub>n</sub>} ，取出若干数组成新的序列 {S<sub>i1</sub>, S<sub>i2</sub>,..., S<sub>im</sub>}，其中 i1、i2 ... im 保持递增，即新序列中各个数仍然保持原数列中的先后顺序，称新序列为原序列的一个 **子序列** 。
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@ -3072,93 +3324,95 @@ public int maxProfit(int[] prices) {
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}
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```
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### 数组区间
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### 股票交易
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**数组区间和**
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**需要冷却期的股票交易**
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[303. Range Sum Query - Immutable (Easy)](https://leetcode.com/problems/range-sum-query-immutable/description/)
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[309. Best Time to Buy and Sell Stock with Cooldown(Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/description/)
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```html
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Given nums = [-2, 0, 3, -5, 2, -1]
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题目描述：交易之后需要有一天的冷却时间。
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sumRange(0, 2) -> 1
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sumRange(2, 5) -> -1
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sumRange(0, 5) -> -3
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```
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求区间 i \~ j 的和，可以转换为 sum[j] - sum[i-1]，其中 sum[i] 为 0 \~ i 的和。
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<div align="center"> <img src="../pics//a3da4342-078b-43e2-b748-7e71bec50dc4.png"/> </div><br>
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```java
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class NumArray {
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private int[] sums;
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public NumArray(int[] nums) {
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sums = new int[nums.length];
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for (int i = 0; i < nums.length; i++) {
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sums[i] = i == 0 ? nums[0] : sums[i - 1] + nums[i];
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}
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}
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public int sumRange(int i, int j) {
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return i == 0 ? sums[j] : sums[j] - sums[i - 1];
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}
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}
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```
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**子数组最大的和**
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[53. Maximum Subarray (Easy)](https://leetcode.com/problems/maximum-subarray/description/)
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```html
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For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
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the contiguous subarray [4,-1,2,1] has the largest sum = 6.
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```
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```java
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public int maxSubArray(int[] nums) {
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if (nums == null || nums.length == 0) {
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public int maxProfit(int[] prices) {
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if (prices == null || prices.length == 0) {
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return 0;
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}
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int preSum = nums[0];
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int maxSum = preSum;
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for (int i = 1; i < nums.length; i++) {
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preSum = preSum > 0 ? preSum + nums[i] : nums[i];
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maxSum = Math.max(maxSum, preSum);
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int N = prices.length;
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int[] buy = new int[N];
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int[] s1 = new int[N];
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int[] sell = new int[N];
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int[] s2 = new int[N];
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s1[0] = buy[0] = -prices[0];
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sell[0] = s2[0] = 0;
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for (int i = 1; i < N; i++) {
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buy[i] = s2[i - 1] - prices[i];
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s1[i] = Math.max(buy[i - 1], s1[i - 1]);
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sell[i] = Math.max(buy[i - 1], s1[i - 1]) + prices[i];
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s2[i] = Math.max(s2[i - 1], sell[i - 1]);
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}
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return maxSum;
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return Math.max(sell[N - 1], s2[N - 1]);
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}
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```
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**数组中等差递增子区间的个数**
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**需要交易费用的股票交易**
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[413. Arithmetic Slices (Medium)](https://leetcode.com/problems/arithmetic-slices/description/)
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[714. Best Time to Buy and Sell Stock with Transaction Fee (Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/)
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```html
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A = [1, 2, 3, 4]
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return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.
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Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
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Output: 8
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Explanation: The maximum profit can be achieved by:
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Buying at prices[0] = 1
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Selling at prices[3] = 8
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Buying at prices[4] = 4
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Selling at prices[5] = 9
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The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
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```
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dp[i] 表示以 A[i] 为结尾的等差递增子区间的个数。
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题目描述：每交易一次，都要支付一定的费用。
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如果 A[i] - A[i - 1] == A[i - 1] - A[i - 2]，表示 [A[i - 2], A[i - 1], A[i]] 是一个等差递增子区间。如果 [A[i - 3], A[i - 2], A[i - 1]] 是一个等差递增子区间，那么 [A[i - 3], A[i - 2], A[i - 1], A[i]] 也是。因此在这个条件下，dp[i] = dp[i-1] + 1。
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<div align="center"> <img src="../pics//61942711-45a0-4e11-bbc9-434e31436f33.png"/> </div><br>
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```java
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public int numberOfArithmeticSlices(int[] A) {
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if (A == null || A.length == 0) {
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return 0;
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public int maxProfit(int[] prices, int fee) {
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int N = prices.length;
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int[] buy = new int[N];
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int[] s1 = new int[N];
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int[] sell = new int[N];
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int[] s2 = new int[N];
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s1[0] = buy[0] = -prices[0];
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sell[0] = s2[0] = 0;
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for (int i = 1; i < N; i++) {
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buy[i] = Math.max(sell[i - 1], s2[i - 1]) - prices[i];
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s1[i] = Math.max(buy[i - 1], s1[i - 1]);
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sell[i] = Math.max(buy[i - 1], s1[i - 1]) - fee + prices[i];
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s2[i] = Math.max(s2[i - 1], sell[i - 1]);
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}
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int n = A.length;
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int[] dp = new int[n];
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for (int i = 2; i < n; i++) {
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if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
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dp[i] = dp[i - 1] + 1;
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return Math.max(sell[N - 1], s2[N - 1]);
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}
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```
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**买入和售出股票最大的收益**
|
||||
|
||||
[121. Best Time to Buy and Sell Stock (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
|
||||
|
||||
题目描述：只进行一次交易。
|
||||
|
||||
只要记录前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。
|
||||
|
||||
```java
|
||||
public int maxProfit(int[] prices) {
|
||||
int n = prices.length;
|
||||
if (n == 0) return 0;
|
||||
int soFarMin = prices[0];
|
||||
int max = 0;
|
||||
for (int i = 1; i < n; i++) {
|
||||
if (soFarMin > prices[i]) soFarMin = prices[i];
|
||||
else max = Math.max(max, prices[i] - soFarMin);
|
||||
}
|
||||
int total = 0;
|
||||
for (int cnt : dp) {
|
||||
total += cnt;
|
||||
}
|
||||
return total;
|
||||
return max;
|
||||
}
|
||||
```
|
||||
|
||||
|
|
@ -3249,257 +3503,6 @@ public int minDistance(String word1, String word2) {
|
|||
}
|
||||
```
|
||||
|
||||
### 分割整数
|
||||
|
||||
**分割整数的最大乘积**
|
||||
|
||||
[343. Integer Break (Medim)](https://leetcode.com/problems/integer-break/description/)
|
||||
|
||||
题目描述：For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).
|
||||
|
||||
```java
|
||||
public int integerBreak(int n) {
|
||||
int[] dp = new int[n + 1];
|
||||
dp[1] = 1;
|
||||
for (int i = 2; i <= n; i++) {
|
||||
for (int j = 1; j <= i - 1; j++) {
|
||||
dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j)));
|
||||
}
|
||||
}
|
||||
return dp[n];
|
||||
}
|
||||
```
|
||||
|
||||
**按平方数来分割整数**
|
||||
|
||||
[279. Perfect Squares(Medium)](https://leetcode.com/problems/perfect-squares/description/)
|
||||
|
||||
题目描述：For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
|
||||
|
||||
```java
|
||||
public int numSquares(int n) {
|
||||
List<Integer> squareList = generateSquareList(n);
|
||||
int[] dp = new int[n + 1];
|
||||
for (int i = 1; i <= n; i++) {
|
||||
int min = Integer.MAX_VALUE;
|
||||
for (int square : squareList) {
|
||||
if (square > i) {
|
||||
break;
|
||||
}
|
||||
min = Math.min(min, dp[i - square] + 1);
|
||||
}
|
||||
dp[i] = min;
|
||||
}
|
||||
return dp[n];
|
||||
}
|
||||
|
||||
private List<Integer> generateSquareList(int n) {
|
||||
List<Integer> squareList = new ArrayList<>();
|
||||
int diff = 3;
|
||||
int square = 1;
|
||||
while (square <= n) {
|
||||
squareList.add(square);
|
||||
square += diff;
|
||||
diff += 2;
|
||||
}
|
||||
return squareList;
|
||||
}
|
||||
```
|
||||
|
||||
**分割整数构成字母字符串**
|
||||
|
||||
[91. Decode Ways (Medium)](https://leetcode.com/problems/decode-ways/description/)
|
||||
|
||||
题目描述：Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12).
|
||||
|
||||
```java
|
||||
public int numDecodings(String s) {
|
||||
if (s == null || s.length() == 0) {
|
||||
return 0;
|
||||
}
|
||||
int n = s.length();
|
||||
int[] dp = new int[n + 1];
|
||||
dp[0] = 1;
|
||||
dp[1] = s.charAt(0) == '0' ? 0 : 1;
|
||||
for (int i = 2; i <= n; i++) {
|
||||
int one = Integer.valueOf(s.substring(i - 1, i));
|
||||
if (one != 0) {
|
||||
dp[i] += dp[i - 1];
|
||||
}
|
||||
if (s.charAt(i - 2) == '0') {
|
||||
continue;
|
||||
}
|
||||
int two = Integer.valueOf(s.substring(i - 2, i));
|
||||
if (two <= 26) {
|
||||
dp[i] += dp[i - 2];
|
||||
}
|
||||
}
|
||||
return dp[n];
|
||||
}
|
||||
```
|
||||
|
||||
### 矩阵路径
|
||||
|
||||
**矩阵的总路径数**
|
||||
|
||||
[62. Unique Paths (Medium)](https://leetcode.com/problems/unique-paths/description/)
|
||||
|
||||
题目描述：统计从矩阵左上角到右下角的路径总数，每次只能向右或者向下移动。
|
||||
|
||||
<div align="center"> <img src="../pics//7c98e1b6-c446-4cde-8513-5c11b9f52aea.jpg"/> </div><br>
|
||||
|
||||
```java
|
||||
public int uniquePaths(int m, int n) {
|
||||
int[] dp = new int[n];
|
||||
Arrays.fill(dp, 1);
|
||||
for (int i = 1; i < m; i++) {
|
||||
for (int j = 1; j < n; j++) {
|
||||
dp[j] = dp[j] + dp[j - 1];
|
||||
}
|
||||
}
|
||||
return dp[n - 1];
|
||||
}
|
||||
```
|
||||
|
||||
也可以直接用数学公式求解，这是一个组合问题。机器人总共移动的次数 S=m+n-2，向下移动的次数 D=m-1，那么问题可以看成从 S 从取出 D 个位置的组合数量，这个问题的解为 C(S, D)。
|
||||
|
||||
```java
|
||||
public int uniquePaths(int m, int n) {
|
||||
int S = m + n - 2; // 总共的移动次数
|
||||
int D = m - 1; // 向下的移动次数
|
||||
long ret = 1;
|
||||
for (int i = 1; i <= D; i++) {
|
||||
ret = ret * (S - D + i) / i;
|
||||
}
|
||||
return (int) ret;
|
||||
}
|
||||
```
|
||||
|
||||
**矩阵的最小路径和**
|
||||
|
||||
[64. Minimum Path Sum (Medium)](https://leetcode.com/problems/minimum-path-sum/description/)
|
||||
|
||||
```html
|
||||
[[1,3,1],
|
||||
[1,5,1],
|
||||
[4,2,1]]
|
||||
Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.
|
||||
```
|
||||
|
||||
题目描述：求从矩阵的左上角到右下角的最小路径和，每次只能向左和向下移动。
|
||||
|
||||
```java
|
||||
public int minPathSum(int[][] grid) {
|
||||
if (grid.length == 0 || grid[0].length == 0) {
|
||||
return 0;
|
||||
}
|
||||
int m = grid.length, n = grid[0].length;
|
||||
int[] dp = new int[n];
|
||||
for (int i = 0; i < m; i++) {
|
||||
for (int j = 0; j < n; j++) {
|
||||
if (j == 0) {
|
||||
dp[0] = dp[0] + grid[i][0]; // 只能从上侧走到该位置
|
||||
} else if (i == 0) {
|
||||
dp[j] = dp[j - 1] + grid[0][j]; // 只能从右侧走到该位置
|
||||
} else {
|
||||
dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j];
|
||||
}
|
||||
}
|
||||
}
|
||||
return dp[n - 1];
|
||||
}
|
||||
```
|
||||
|
||||
### 其它问题
|
||||
|
||||
**需要冷却期的股票交易**
|
||||
|
||||
[309. Best Time to Buy and Sell Stock with Cooldown(Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/description/)
|
||||
|
||||
题目描述：交易之后需要有一天的冷却时间。
|
||||
|
||||
<div align="center"> <img src="../pics//a3da4342-078b-43e2-b748-7e71bec50dc4.png"/> </div><br>
|
||||
|
||||
```java
|
||||
public int maxProfit(int[] prices) {
|
||||
if (prices == null || prices.length == 0) return 0;
|
||||
int N = prices.length;
|
||||
int[] buy = new int[N];
|
||||
int[] s1 = new int[N];
|
||||
int[] sell = new int[N];
|
||||
int[] s2 = new int[N];
|
||||
s1[0] = buy[0] = -prices[0];
|
||||
sell[0] = s2[0] = 0;
|
||||
for (int i = 1; i < N; i++) {
|
||||
buy[i] = s2[i - 1] - prices[i];
|
||||
s1[i] = Math.max(buy[i - 1], s1[i - 1]);
|
||||
sell[i] = Math.max(buy[i - 1], s1[i - 1]) + prices[i];
|
||||
s2[i] = Math.max(s2[i - 1], sell[i - 1]);
|
||||
}
|
||||
return Math.max(sell[N - 1], s2[N - 1]);
|
||||
}
|
||||
```
|
||||
|
||||
**需要交易费用的股票交易**
|
||||
|
||||
[714. Best Time to Buy and Sell Stock with Transaction Fee (Medium)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/)
|
||||
|
||||
```html
|
||||
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
|
||||
Output: 8
|
||||
Explanation: The maximum profit can be achieved by:
|
||||
Buying at prices[0] = 1
|
||||
Selling at prices[3] = 8
|
||||
Buying at prices[4] = 4
|
||||
Selling at prices[5] = 9
|
||||
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
|
||||
```
|
||||
|
||||
题目描述：每交易一次，都要支付一定的费用。
|
||||
|
||||
<div align="center"> <img src="../pics//61942711-45a0-4e11-bbc9-434e31436f33.png"/> </div><br>
|
||||
|
||||
```java
|
||||
public int maxProfit(int[] prices, int fee) {
|
||||
int N = prices.length;
|
||||
int[] buy = new int[N];
|
||||
int[] s1 = new int[N];
|
||||
int[] sell = new int[N];
|
||||
int[] s2 = new int[N];
|
||||
s1[0] = buy[0] = -prices[0];
|
||||
sell[0] = s2[0] = 0;
|
||||
for (int i = 1; i < N; i++) {
|
||||
buy[i] = Math.max(sell[i - 1], s2[i - 1]) - prices[i];
|
||||
s1[i] = Math.max(buy[i - 1], s1[i - 1]);
|
||||
sell[i] = Math.max(buy[i - 1], s1[i - 1]) - fee + prices[i];
|
||||
s2[i] = Math.max(s2[i - 1], sell[i - 1]);
|
||||
}
|
||||
return Math.max(sell[N - 1], s2[N - 1]);
|
||||
}
|
||||
```
|
||||
|
||||
**买入和售出股票最大的收益**
|
||||
|
||||
[121. Best Time to Buy and Sell Stock (Easy)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
|
||||
|
||||
只进行一次交易。
|
||||
|
||||
只要记录前面的最小价格，将这个最小价格作为买入价格，然后将当前的价格作为售出价格，查看当前收益是不是最大收益。
|
||||
|
||||
```java
|
||||
public int maxProfit(int[] prices) {
|
||||
int n = prices.length;
|
||||
if(n == 0) return 0;
|
||||
int soFarMin = prices[0];
|
||||
int max = 0;
|
||||
for(int i = 1; i < n; i++){
|
||||
if(soFarMin > prices[i]) soFarMin = prices[i];
|
||||
else max = Math.max(max, prices[i] - soFarMin);
|
||||
}
|
||||
return max;
|
||||
}
|
||||
```
|
||||
|
||||
**复制粘贴字符**
|
||||
|
||||
[650. 2 Keys Keyboard (Medium)](https://leetcode.com/problems/2-keys-keyboard/description/)
|
||||
|
|
@ -3516,12 +3519,23 @@ In step 2, we use Paste operation to get 'AA'.
|
|||
In step 3, we use Paste operation to get 'AAA'.
|
||||
```
|
||||
|
||||
```java
|
||||
public int minSteps(int n) {
|
||||
if (n == 1) return 0;
|
||||
for (int i = 2; i <= Math.sqrt(n); i++) {
|
||||
if (n % i == 0) return i + minSteps(n / i);
|
||||
}
|
||||
return n;
|
||||
}
|
||||
```
|
||||
|
||||
```java
|
||||
public int minSteps(int n) {
|
||||
int[] dp = new int[n + 1];
|
||||
int h = (int) Math.sqrt(n);
|
||||
for (int i = 2; i <= n; i++) {
|
||||
dp[i] = i;
|
||||
for (int j = i - 1; j >= 0; j--) {
|
||||
for (int j = 2; j <= h; j++) {
|
||||
if (i % j == 0) {
|
||||
dp[i] = dp[j] + dp[i / j];
|
||||
break;
|
||||
|
|
@ -3532,16 +3546,6 @@ public int minSteps(int n) {
|
|||
}
|
||||
```
|
||||
|
||||
```java
|
||||
public int minSteps(int n) {
|
||||
if (n == 1) return 0;
|
||||
for (int i = 2; i <= Math.sqrt(n); i++) {
|
||||
if (n % i == 0) return i + minSteps(n / i);
|
||||
}
|
||||
return n;
|
||||
}
|
||||
```
|
||||
|
||||
## 数学
|
||||
|
||||
### 素数
|
||||
|
|
@ -6660,8 +6664,8 @@ x ^ x = 0 x & x = x x | x = x
|
|||
位与运算技巧：
|
||||
|
||||
- n&(n-1) 去除 n 的位级表示中最低的那一位。例如对于二进制表示 10110 **100** ，减去 1 得到 10110**011**，这两个数相与得到 10110**000**。
|
||||
- n&(-n) 得到 n 的位级表示中最低的那一位。-n 得到 n 的反码加 1，对于二进制表示 10110 **100** ，-n 得到 01001**100**，相与得到 00000**100**。
|
||||
- n-n&(\~n+1) 去除 n 的位级表示中最高的那一位。
|
||||
- n&(-n) 得到 n 的位级表示中最低的那一位。-n 得到 n 的反码加 1，对于二进制表示 10110 **100** ，-n 得到 01001**100**，相与得到 00000**100**
|
||||
|
||||
移位运算：
|
||||
|
||||
|
|
@ -6794,8 +6798,7 @@ diff &= -diff 得到出 diff 最右侧不为 0 的位，也就是不存在重复
|
|||
public int[] singleNumber(int[] nums) {
|
||||
int diff = 0;
|
||||
for (int num : nums) diff ^= num;
|
||||
// 得到最右一位
|
||||
diff &= -diff;
|
||||
diff &= -diff; // 得到最右一位
|
||||
int[] ret = new int[2];
|
||||
for (int num : nums) {
|
||||
if ((num & diff) == 0) ret[0] ^= num;
|
||||
|
|
@ -6829,13 +6832,11 @@ private static Map<Byte, Integer> cache = new HashMap<>();
|
|||
public int reverseBits(int n) {
|
||||
int ret = 0;
|
||||
for (int i = 0; i < 4; i++) {
|
||||
byte b = (byte) (n & 0b11111111);
|
||||
ret <<= 8;
|
||||
ret |= reverseByte(b);
|
||||
ret |= reverseByte((byte) (n & 0b11111111));
|
||||
n >>= 8;
|
||||
}
|
||||
return ret;
|
||||
|
||||
}
|
||||
|
||||
private int reverseByte(byte b) {
|
||||
|
|
|
|||
Loading…
Reference in New Issue
Block a user