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CyC2018
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@ -13,14 +13,14 @@
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* [Backtracking](#backtracking)
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* [分治](#分治)
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* [动态规划](#动态规划)
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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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* [斐波那契数列](#斐波那契数列)
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* [最长递增子序列](#最长递增子序列)
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* [最长公共子系列](#最长公共子系列)
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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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* [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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@ -1719,130 +1719,6 @@ public List<Integer> diffWaysToCompute(String input) {
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递归和动态规划都是将原问题拆成多个子问题然后求解，他们之间最本质的区别是，动态规划保存了子问题的解。
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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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[Leetcode : 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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[Leetcode : 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 max = Integer.MAX_VALUE;
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for (int square : squareList) {
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if (square > i) break;
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max = Math.min(max, dp[i - square] + 1);
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}
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dp[i] = max;
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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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[Leetcode : 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) return 0;
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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) dp[i] += dp[i - 1];
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if(s.charAt(i - 2) == '0') continue;
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int two = Integer.valueOf(s.substring(i - 2, i));
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if(two <= 26) dp[i] += dp[i - 2];
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}
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return dp[n];
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}
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```
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### 矩阵路径
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**矩阵的总路径数**
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[Leetcode : 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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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if(i == 0) dp[j] = 1;
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else if(j != 0) 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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**矩阵的最小路径和**
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[Leetcode : 64. Minimum Path Sum (Medium)](https://leetcode.com/problems/minimum-path-sum/description/)
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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，每次只能向左和向下移动。
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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) return 0;
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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) dp[0] = dp[0] + grid[i][0];
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else if(i == 0) dp[j] = dp[j - 1] + grid[0][j];
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else dp[j] = Math.min(dp[j - 1], 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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### 斐波那契数列
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**爬楼梯**
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@ -1863,7 +1739,6 @@ dp[N] 即为所求。
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public int climbStairs(int n) {
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if(n == 1) return 1;
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if(n == 2) return 2;
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// 前一个楼梯、后一个楼梯
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int pre1 = 2, pre2 = 1;
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for(int i = 2; i < n; i++){
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int cur = pre1 + pre2;
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@ -1937,31 +1812,32 @@ public int rob(int[] nums) {
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[Leetcode : 213. House Robber II (Medium)](https://leetcode.com/problems/house-robber-ii/description/)
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```java
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private int[] dp;
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public int rob(int[] nums) {
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if (nums == null || nums.length == 0) return 0;
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int n = nums.length;
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if (n == 1) return nums[0];
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dp = new int[n];
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return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
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}
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private int rob(int[] nums, int s, int e) {
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int n = nums.length;
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if(e - s == 0) return nums[s];
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if(e - s == 1) return Math.max(nums[s], nums[s + 1]);
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int[] dp = new int[n];
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dp[s] = nums[s];
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dp[s + 1] = nums[s + 1];
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dp[s + 2] = nums[s] + nums[s + 2];
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for (int i = s + 3; i <= e; i++) {
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private int rob(int[] nums, int first, int last) {
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if (last - first == 0) return nums[first];
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if (last - first == 1) return Math.max(nums[first], nums[first + 1]);
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dp[first] = nums[first];
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dp[first + 1] = nums[first + 1];
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dp[first + 2] = nums[first] + nums[first + 2];
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for (int i = first + 3; i <= last; i++) {
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dp[i] = Math.max(dp[i - 2], dp[i - 3]) + nums[i];
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}
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return Math.max(dp[e], dp[e - 1]);
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return Math.max(dp[last], dp[last - 1]);
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}
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```
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**信件错排**
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题目描述：有 N 个 信 和 信封，它们被打乱，求错误装信的方式数量。
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题目描述：有 N 个 信 和 信封，它们被打乱，求错误装信方式的数量。
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定义一个数组 dp 存储错误方式数量，dp[i] 表示前 i 个信和信封的错误方式数量。假设第 i 个信装到第 j 个信封里面，而第 j 个信装到第 k 个信封里面。根据 i 和 k 是否相等，有两种情况：
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@ -2545,6 +2421,137 @@ public int minDistance(String word1, String word2) {
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[Leetcode : 72. Edit Distance (Hard)](https://leetcode.com/problems/edit-distance/description/)
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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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[Leetcode : 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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[Leetcode : 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 max = Integer.MAX_VALUE;
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for (int square : squareList) {
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if (square > i) break;
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max = Math.min(max, dp[i - square] + 1);
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}
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dp[i] = max;
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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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[Leetcode : 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) return 0;
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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) dp[i] += dp[i - 1];
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if(s.charAt(i - 2) == '0') continue;
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int two = Integer.valueOf(s.substring(i - 2, i));
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if(two <= 26) dp[i] += dp[i - 2];
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}
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return dp[n];
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}
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```
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### 矩阵路径
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**矩阵的总路径数**
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[Leetcode : 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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**矩阵的最小路径和**
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[Leetcode : 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，每次只能向左和向下移动。
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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) return 0;
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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) dp[0] = dp[0] + grid[i][0]; // 只能从上侧走到该位置
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else if (i == 0) dp[j] = dp[j - 1] + grid[0][j]; // 只能从右侧走到该位置
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else dp[j] = Math.min(dp[j - 1], 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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### 其它问题
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**需要冷却期的股票交易**
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