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CyC2018
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@ -1890,12 +1890,12 @@ public int lengthOfLIS(int[] nums) {
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}
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```
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以上解法的时间复杂度为 O(n<sup>2</sup>) ，可以使用二分查找使得时间复杂度降低为 O(nlog<sub>n</sub>)。定义一个 tails 数组，其中 tails[i] 存储长度为 i + 1 的最长递增子序列的最后一个元素，例如对于数组 [4,5,6,3]，有
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以上解法的时间复杂度为 O(n<sup>2</sup>) ，可以使用二分查找使得时间复杂度降低为 O(nlogn)。定义一个 tails 数组，其中 tails[i] 存储长度为 i + 1 的最长递增子序列的最后一个元素，例如对于数组 [4,5,6,3]，有
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```html
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len = 1 : [4], [5], [6], [3] => tails[0] = 3
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len = 2 : [4, 5], [5, 6] => tails[1] = 5
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len = 3 : [4, 5, 6] => tails[2] = 6
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len = 3 : [4, 5, 6] => tails[2] = 6
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```
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对于一个元素 x，如果它大于 tails 数组所有的值，那么把它添加到 tails 后面；如果 tails[i-1] < x <= tails[i]，那么更新 tails[i] = x 。
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@ -1907,22 +1907,22 @@ public int lengthOfLIS(int[] nums) {
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int n = nums.length;
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int[] tails = new int[n];
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int size = 0;
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for(int i = 0; i < n; i++){
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int idx = binarySearch(tails, 0, size, nums[i]);
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tails[idx] = nums[i];
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if(idx == size) size++;
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for (int i = 0; i < n; i++) {
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int index = binarySearch(tails, 0, size, nums[i]);
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tails[index] = nums[i];
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if (index == size) size++;
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}
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return size;
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}
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private int binarySearch(int[] nums, int sIdx, int eIdx, int key){
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while(sIdx < eIdx){
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int mIdx = sIdx + (eIdx - sIdx) / 2;
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if(nums[mIdx] == key) return mIdx;
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else if(nums[mIdx] > key) eIdx = mIdx;
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else sIdx = mIdx + 1;
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private int binarySearch(int[] nums, int first, int last, int key) {
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while (first < last) {
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int mid = first + (last - first) / 2;
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if (nums[mid] == key) return mid;
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else if (nums[mid] > key) last = mid;
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else first = mid + 1;
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}
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return sIdx;
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return first;
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}
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```
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@ -1930,6 +1930,19 @@ private int binarySearch(int[] nums, int sIdx, int eIdx, int key){
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[Leetcode : 376. Wiggle Subsequence (Medium)](https://leetcode.com/problems/wiggle-subsequence/description/)
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```html
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Input: [1,7,4,9,2,5]
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Output: 6
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The entire sequence is a wiggle sequence.
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Input: [1,17,5,10,13,15,10,5,16,8]
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Output: 7
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There are several subsequences that achieve this length. One is [1,17,10,13,10,16,8].
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Input: [1,2,3,4,5,6,7,8,9]
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Output: 2
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```
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要求：使用 O(n) 时间复杂度求解。
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使用两个状态 up 和 down。
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@ -2008,7 +2021,7 @@ public int knapsack(int W, int N, int[] weights, int[] values) {
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for (int j = W - 1; j >= weights[i]; j--) {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weights[i]] + values[i]);
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}
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for (int j = weights[i - 1] - 1; j >= 0; j--) {
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for (int j = weights[i] - 1; j >= 0; j--) {
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dp[i][j] = dp[i - 1][j];
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}
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}
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@ -2048,6 +2061,14 @@ public int knapsack(int W, int N, int[] weights, int[] values) {
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[Leetcode : 416. Partition Equal Subset Sum (Medium)](https://leetcode.com/problems/partition-equal-subset-sum/description/)
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```html
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Input: [1, 5, 11, 5]
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Output: true
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Explanation: The array can be partitioned as [1, 5, 5] and [11].
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```
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可以看成一个背包大小为 sum/2 的 0-1 背包问题，但是也有不同的地方，这里没有价值属性，并且背包必须被填满。
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以下实现使用了空间优化。
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@ -2063,16 +2084,16 @@ public boolean canPartition(int[] nums) {
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}
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int W = sum / 2;
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boolean[] dp = new boolean[W + 1];
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int n = nums.length;
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for(int i = 0; i <= W; i++) {
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if(nums[0] == i) dp[i] = true;
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for (int i = 0; i <= W; i++) {
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if (nums[0] == i) {
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dp[i] = true;
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}
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}
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for(int i = 1; i < n; i++) {
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for(int j = W; j >= nums[i]; j--) {
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for (int i = 1; i < nums.length; i++) {
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for (int j = W; j >= nums[i]; j--) {
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dp[j] = dp[j] || dp[j - nums[i]];
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}
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}
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return dp[W];
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}
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```
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@ -2094,8 +2115,7 @@ public boolean wordBreak(String s, List<String> wordDict) {
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dp[0] = true;
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for (int i = 1; i <= n; i++) {
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for (String word : wordDict) {
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if (word.length() <= i
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&& word.equals(s.substring(i - word.length(), i))) {
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if (word.length() <= i && word.equals(s.substring(i - word.length(), i))) {
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dp[i] = dp[i] || dp[i - word.length()];
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}
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}
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@ -2174,9 +2194,8 @@ Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3
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```java
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public int findMaxForm(String[] strs, int m, int n) {
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if (strs == null || strs.length == 0) return 0;
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int l = strs.length;
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int[][] dp = new int[m + 1][n + 1];
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for (int i = 0; i < l; i++) {
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for (int i = 0; i < strs.length; i++) {
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String s = strs[i];
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int ones = 0, zeros = 0;
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for (char c : s.toCharArray()) {
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@ -2185,9 +2204,7 @@ public int findMaxForm(String[] strs, int m, int n) {
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}
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for (int j = m; j >= zeros; j--) {
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for (int k = n; k >= ones; k--) {
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if (zeros <= j && ones <= k) {
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dp[j][k] = Math.max(dp[j][k], dp[j - zeros][k - ones] + 1);
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}
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dp[j][k] = Math.max(dp[j][k], dp[j - zeros][k - ones] + 1);
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}
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}
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}
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@ -2199,20 +2216,30 @@ public int findMaxForm(String[] strs, int m, int n) {
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[Leetcode : 322. Coin Change (Medium)](https://leetcode.com/problems/coin-change/description/)
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```html
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Example 1:
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coins = [1, 2, 5], amount = 11
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return 3 (11 = 5 + 5 + 1)
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Example 2:
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coins = [2], amount = 3
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return -1.
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```
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题目描述：给一些面额的硬币，要求用这些硬币来组成给定面额的钱数，并且使得硬币数量最少。硬币可以重复使用。
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这是一个完全背包问题，完全背包问题和 0-1 背包问题在实现上唯一的不同是，第二层循环是从 0 开始的，而不是从尾部开始。
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```java
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public int coinChange(int[] coins, int amount) {
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if (coins == null || coins.length == 0) return 0;
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Arrays.sort(coins);
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int[] dp = new int[amount + 1];
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Arrays.fill(dp, amount + 1);
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dp[0] = 0;
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for (int i = 1; i <= amount; i++) {
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for (int j = 0; j < coins.length; j++) {
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if (coins[j] <= i) {
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dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
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}
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for (int j = 0; j < coins.length && coins[j] <= i; j++) {
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dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
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}
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}
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return dp[amount] > amount ? -1 : dp[amount];
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@ -2243,11 +2270,12 @@ Therefore the output is 7.
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```java
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public int combinationSum4(int[] nums, int target) {
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if (nums == null || nums.length == 0) return 0;
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int[] dp = new int[target + 1];
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dp[0] = 1;
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for (int i = 1; i <= target; i++) {
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for (int j = 0; j < nums.length; j++) {
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if(nums[j] <= i) {
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if (nums[j] <= i) {
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dp[i] += dp[i - nums[j]];
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}
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}
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@ -2285,7 +2313,7 @@ dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range o
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```java
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public int maxProfit(int k, int[] prices) {
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int n = prices.length;
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if (k >= n/2) {
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if (k >= n/2) { // 这种情况下该问题退化为普通的股票交易问题
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int maxPro = 0;
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for (int i = 1; i < n; i++) {
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if (prices[i] > prices[i-1])
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@ -975,7 +975,7 @@ public class BinarySearchST<Key extends Comparable<Key>, Value> {
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<div align="center"> <img src="../pics//1c012d74-6b9d-4f25-a016-7ad4f1f1521898780376.png" width="400"/> </div><br>
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BST 有一个重要性质，就是它的前序遍历结果递增排序。
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BST 有一个重要性质，就是它的中序遍历结果递增排序。
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<div align="center"> <img src="../pics//5c0bb285-b917-446b-84a2-9810ee41521898714517.png" width="300"/> </div><br>
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