diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md index 4aab1782..8925ef81 100644 --- a/problems/0040.组合总和II.md +++ b/problems/0040.组合总和II.md @@ -392,7 +392,66 @@ var combinationSum2 = function(candidates, target) { } }; ``` +C: +```c +int* path; +int pathTop; +int** ans; +int ansTop; +//记录ans中每一个一维数组的大小 +int* length; +int cmp(const void* a1, const void* a2) { + return *((int*)a1) - *((int*)a2); +} +void backTracking(int* candidates, int candidatesSize, int target, int sum, int startIndex) { + if(sum >= target) { + //若sum等于target，复制当前path进入 + if(sum == target) { + int* tempPath = (int*)malloc(sizeof(int) * pathTop); + int j; + for(j = 0; j < pathTop; j++) { + tempPath[j] = path[j]; + } + length[ansTop] = pathTop; + ans[ansTop++] = tempPath; + } + return ; + } + + int i; + for(i = startIndex; i < candidatesSize; i++) { + //对同一层树中使用过的元素跳过 + if(i > startIndex && candidates[i] == candidates[i-1]) + continue; + path[pathTop++] = candidates[i]; + sum += candidates[i]; + backTracking(candidates, candidatesSize, target, sum, i + 1); + //回溯 + sum -= candidates[i]; + pathTop--; + } +} + +int** combinationSum2(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes){ + path = (int*)malloc(sizeof(int) * 50); + ans = (int**)malloc(sizeof(int*) * 100); + length = (int*)malloc(sizeof(int) * 100); + pathTop = ansTop = 0; + //快速排序candidates，让相同元素挨到一起 + qsort(candidates, candidatesSize, sizeof(int), cmp); + + backTracking(candidates, candidatesSize, target, 0, 0); + + *returnSize = ansTop; + *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop); + int i; + for(i = 0; i < ansTop; i++) { + (*returnColumnSizes)[i] = length[i]; + } + return ans; +} +``` ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 6386c48d..e8a98527 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -408,25 +408,57 @@ class Solution { Python： ```Python class Solution: + """二叉树的所有路径 递归法""" + def binaryTreePaths(self, root: TreeNode) -> List[str]: - path=[] - res=[] - def backtrace(root, path): - if not root:return - path.append(root.val) - if (not root.left)and (not root.right): - res.append(path[:]) - ways=[] - if root.left:ways.append(root.left) - if root.right:ways.append(root.right) - for way in ways: - backtrace(way,path) - path.pop() - backtrace(root,path) - return ["->".join(list(map(str,i))) for i in res] + path, result = '', [] + self.traversal(root, path, result) + return result + + def traversal(self, cur: TreeNode, path: List, result: List): + path += str(cur.val) + # 如果当前节点为叶子节点，添加路径到结果中 + if not (cur.left or cur.right): + result.append(path) + return + + if cur.left: + self.traversal(cur.left, path + '->', result) + + if cur.right: + self.traversal(cur.right, path + '->', result) ``` +```python +from collections import deque + + +class Solution: + """二叉树的所有路径 迭代法""" + + def binaryTreePaths(self, root: TreeNode) -> List[str]: + # 题目中节点数至少为1 + stack, path_st, result = deque([root]), deque(), [] + path_st.append(str(root.val)) + + while stack: + cur = stack.pop() + path = path_st.pop() + # 如果当前节点为叶子节点，添加路径到结果中 + if not (cur.left or cur.right): + result.append(path) + if cur.right: + stack.append(cur.right) + path_st.append(path + '->' + str(cur.right.val)) + if cur.left: + stack.append(cur.left) + path_st.append(path + '->' + str(cur.left.val)) + + return result +``` + + Go： ```go func binaryTreePaths(root *TreeNode) []string { diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 4993bede..07cf0433 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -321,7 +321,7 @@ const findTargetSumWays = (nums, target) => { const sum = nums.reduce((a, b) => a+b); - if(target > sum) { + if(Math.abs(target) > sum) { return 0; } diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 1a6d39af..b644a0a3 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -258,16 +258,30 @@ class Solution { ## Python ```python -//递归法 class Solution: + """最大二叉树 递归法""" + def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: - if not nums: return None //终止条件 - root = TreeNode(max(nums)) //新建节点 - p = nums.index(root.val) //找到最大值位置 - if p > 0: //保证有左子树 - root.left = self.constructMaximumBinaryTree(nums[:p]) //递归 - if p < len(nums): //保证有右子树 - root.right = self.constructMaximumBinaryTree(nums[p+1:]) //递归 + return self.traversal(nums, 0, len(nums)) + + def traversal(self, nums: List[int], begin: int, end: int) -> TreeNode: + # 列表长度为0时返回空节点 + if begin == end: + return None + + # 找到最大的值和其对应的下标 + max_index = begin + for i in range(begin, end): + if nums[i] > nums[max_index]: + max_index = i + + # 构建当前节点 + root = TreeNode(nums[max_index]) + + # 递归构建左右子树 + root.left = self.traversal(nums, begin, max_index) + root.right = self.traversal(nums, max_index + 1, end) + return root ``` diff --git a/problems/为了绝杀编辑距离，卡尔做了三步铺垫.md b/problems/为了绝杀编辑距离，卡尔做了三步铺垫.md index 31a5e448..2eb253ba 100644 --- a/problems/为了绝杀编辑距离，卡尔做了三步铺垫.md +++ b/problems/为了绝杀编辑距离，卡尔做了三步铺垫.md @@ -167,7 +167,33 @@ else { Java： - +```java +class Solution { + public int minDistance(String word1, String word2) { + int m = word1.length(); + int n = word2.length(); + int[][] dp = new int[m+1][n+1]; + for(int i = 1; i <= m; i++){ + dp[i][0] = i; + } + for(int i = 1; i <= n; i++){ + dp[0][i] = i; + } + for(int i = 1; i <= m; i++){ + for(int j = 1; j <= n; j++){ + int left = dp[i][j-1]+1; + int mid = dp[i-1][j-1]; + int right = dp[i-1][j]+1; + if(word1.charAt(i-1) != word2.charAt(j-1)){ + mid ++; + } + dp[i][j] = Math.min(left,Math.min(mid,right)); + } + } + return dp[m][n]; + } +} +``` Python： diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index 2bcded70..7d04b514 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -34,7 +34,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目， ## 01 背包 -有N件物品和一个最多能被重量为W 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。**每件物品只能用一次**，求解将哪些物品装入背包里物品价值总和最大。 +有N件物品和一个最多能背重量为W 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。**每件物品只能用一次**，求解将哪些物品装入背包里物品价值总和最大。 ![动态规划-背包问题](https://img-blog.csdnimg.cn/20210117175428387.jpg)