diff --git "a/problems/0001.\344\270\244\346\225\260\344\271\213\345\222\214.md" "b/problems/0001.\344\270\244\346\225\260\344\271\213\345\222\214.md" index e3fb0fb55f..bf1e173ec6 100644 --- "a/problems/0001.\344\270\244\346\225\260\344\271\213\345\222\214.md" +++ "b/problems/0001.\344\270\244\346\225\260\344\271\213\345\222\214.md" @@ -163,7 +163,7 @@ class Solution: for index, value in enumerate(nums): if target - value in records: # 遍历当前元素，并在map中寻找是否有匹配的key return [records[target- value], index] - records[value] = index # 遍历当前元素，并在map中寻找是否有匹配的key + records[value] = index # 如果没找到匹配对，就把访问过的元素和下标加入到map中 return [] ``` （版本二）使用集合 diff --git "a/problems/0077.\347\273\204\345\220\210.md" "b/problems/0077.\347\273\204\345\220\210.md" index 7d71b51763..8d448739c0 100644 --- "a/problems/0077.\347\273\204\345\220\210.md" +++ "b/problems/0077.\347\273\204\345\220\210.md" @@ -343,8 +343,32 @@ public: ## 其他语言版本 -### Java +### Java： +未剪枝优化 +```java +class Solution { + List> result= new ArrayList<>(); + LinkedList path = new LinkedList<>(); + public List> combine(int n, int k) { + backtracking(n,k,1); + return result; + } + + public void backtracking(int n,int k,int startIndex){ + if (path.size() == k){ + result.add(new ArrayList<>(path)); + return; + } + for (int i =startIndex;i<=n;i++){ + path.add(i); + backtracking(n,k,i+1); + path.removeLast(); + } + } +} +``` +剪枝优化： ```java class Solution { List> result = new ArrayList<>(); diff --git "a/problems/0450.\345\210\240\351\231\244\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\350\212\202\347\202\271.md" "b/problems/0450.\345\210\240\351\231\244\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\350\212\202\347\202\271.md" index 13599fd849..18e5cb4cb1 100644 --- "a/problems/0450.\345\210\240\351\231\244\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\350\212\202\347\202\271.md" +++ "b/problems/0450.\345\210\240\351\231\244\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\350\212\202\347\202\271.md" @@ -268,6 +268,34 @@ public: ### Java +```java +// 解法1(最好理解的版本) +class Solution { + public TreeNode deleteNode(TreeNode root, int key) { + if (root == null) return root; + if (root.val == key) { + if (root.left == null) { + return root.right; + } else if (root.right == null) { + return root.left; + } else { + TreeNode cur = root.right; + while (cur.left != null) { + cur = cur.left; + } + cur.left = root.left; + root = root.right; + return root; + } + } + if (root.val > key) root.left = deleteNode(root.left, key); + if (root.val < key) root.right = deleteNode(root.right, key); + return root; + } +} +``` + + ```java class Solution { public TreeNode deleteNode(TreeNode root, int key) { @@ -296,34 +324,59 @@ class Solution { } } ``` +递归法 ```java -// 解法2 class Solution { public TreeNode deleteNode(TreeNode root, int key) { - if (root == null) return root; - if (root.val == key) { - if (root.left == null) { - return root.right; - } else if (root.right == null) { - return root.left; - } else { - TreeNode cur = root.right; - while (cur.left != null) { - cur = cur.left; - } - cur.left = root.left; - root = root.right; - return root; + if (root == null){ + return null; + } + //寻找对应的对应的前面的节点，以及他的前一个节点 + TreeNode cur = root; + TreeNode pre = null; + while (cur != null){ + if (cur.val < key){ + pre = cur; + cur = cur.right; + } else if (cur.val > key) { + pre = cur; + cur = cur.left; + }else { + break; } } - if (root.val > key) root.left = deleteNode(root.left, key); - if (root.val < key) root.right = deleteNode(root.right, key); + if (pre == null){ + return deleteOneNode(cur); + } + if (pre.left !=null && pre.left.val == key){ + pre.left = deleteOneNode(cur); + } + if (pre.right !=null && pre.right.val == key){ + pre.right = deleteOneNode(cur); + } return root; } + + public TreeNode deleteOneNode(TreeNode node){ + if (node == null){ + return null; + } + if (node.right == null){ + return node.left; + } + TreeNode cur = node.right; + while (cur.left !=null){ + cur = cur.left; + } + cur.left = node.left; + return node.right; + } } ``` + ### Python + 递归法（版本一） ```python class Solution: diff --git "a/problems/0496.\344\270\213\344\270\200\344\270\252\346\233\264\345\244\247\345\205\203\347\264\240I.md" "b/problems/0496.\344\270\213\344\270\200\344\270\252\346\233\264\345\244\247\345\205\203\347\264\240I.md" index 31c3ce4387..411a47df9b 100644 --- "a/problems/0496.\344\270\213\344\270\200\344\270\252\346\233\264\345\244\247\345\205\203\347\264\240I.md" +++ "b/problems/0496.\344\270\213\344\270\200\344\270\252\346\233\264\345\244\247\345\205\203\347\264\240I.md" @@ -387,6 +387,32 @@ function nextGreaterElement(nums1: number[], nums2: number[]): number[] { }; ``` +Rust + +```rust +impl Solution { + pub fn next_greater_element(nums1: Vec, nums2: Vec) -> Vec { + let mut ans = vec![-1; nums1.len()]; + use std::collections::HashMap; + let mut map = HashMap::new(); + for (idx, &i) in nums1.iter().enumerate() { + map.insert(i, idx); + } + let mut stack = vec![]; + for (idx, &i) in nums2.iter().enumerate() { + while !stack.is_empty() && nums2[*stack.last().unwrap()] < i { + let pos = stack.pop().unwrap(); + if let Some(&jdx) = map.get(&nums2[pos]) { + ans[jdx] = i; + } + } + stack.push(idx); + } + ans + } +} +``` +

diff --git "a/problems/0739.\346\257\217\346\227\245\346\270\251\345\272\246.md" "b/problems/0739.\346\257\217\346\227\245\346\270\251\345\272\246.md" index 749dc97235..d2da37371a 100644 --- "a/problems/0739.\346\257\217\346\227\245\346\270\251\345\272\246.md" +++ "b/problems/0739.\346\257\217\346\227\245\346\270\251\345\272\246.md" @@ -455,7 +455,27 @@ function dailyTemperatures(temperatures: number[]): number[] { }; ``` - +Rust: + +```rust +impl Solution { + /// 单调栈的本质是以空间换时间，记录之前已访问过的非递增子序列下标 + pub fn daily_temperatures(temperatures: Vec) -> Vec { + let mut res = vec![0; temperatures.len()]; + let mut stack = vec![]; + for (idx, &value) in temperatures.iter().enumerate() { + while !stack.is_empty() && temperatures[*stack.last().unwrap()] < value { + // 弹出，并计算res中对应位置的值 + let i = stack.pop().unwrap(); + res[i] = (idx - i) as i32; + } + // 入栈 + stack.push(idx) + } + res + } +} +```