diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index 99c870c6..441cc63c 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -176,30 +176,27 @@ class Solution: ``` Go： -```Go +```golang func merge(intervals [][]int) [][]int { - sort.Slice(intervals, func(i, j int) bool { - return intervals[i][0]=intervals[i+1][0]{ + intervals[i][1]=max(intervals[i][1],intervals[i+1][1])//赋值最大值 + intervals=append(intervals[:i+1],intervals[i+2:]...) + i-- + } + } + return intervals } -func max(a, b int) int { - if a > b { return a } - return b +func max(a,b int)int{ + if a>b{ + return a + } + return b } ``` diff --git a/problems/0098.验证二叉搜索树.md b/problems/0098.验证二叉搜索树.md index d6db901e..6f7e5c14 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -103,7 +103,7 @@ if (root->val > root->left->val && root->val < root->right->val) { ![二叉搜索树](https://img-blog.csdnimg.cn/20200812191501419.png) -节点10小于左节点5，大于右节点15，但右子树里出现了一个6 这就不符合了！ +节点10大于左节点5，小于右节点15，但右子树里出现了一个6 这就不符合了！ * 陷阱2 diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index c38cc796..007eac44 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -1046,6 +1046,31 @@ class Solution: out_list.append(max(in_list)) return out_list ``` +java代码： + +```java +class Solution { + public List largestValues(TreeNode root) { + List retVal = new ArrayList(); + Queue tmpQueue = new LinkedList(); + if (root != null) tmpQueue.add(root); + + while (tmpQueue.size() != 0){ + int size = tmpQueue.size(); + List lvlVals = new ArrayList(); + for (int index = 0; index < size; index++){ + TreeNode node = tmpQueue.poll(); + lvlVals.add(node.val); + if (node.left != null) tmpQueue.add(node.left); + if (node.right != null) tmpQueue.add(node.right); + } + retVal.add(Collections.max(lvlVals)); + } + + return retVal; + } +} +``` go: diff --git a/problems/0116.填充每个节点的下一个右侧节点指针.md b/problems/0116.填充每个节点的下一个右侧节点指针.md index f96351aa..34c666f3 100644 --- a/problems/0116.填充每个节点的下一个右侧节点指针.md +++ b/problems/0116.填充每个节点的下一个右侧节点指针.md @@ -130,15 +130,97 @@ public: ## Java ```java - +// 递归法 +class Solution { + public void traversal(Node cur) { + if (cur == null) return; + if (cur.left != null) cur.left.next = cur.right; // 操作1 + if (cur.right != null) { + if(cur.next != null) cur.right.next = cur.next.left; //操作2 + else cur.right.next = null; + } + traversal(cur.left); // 左 + traversal(cur.right); //右 + } + public Node connect(Node root) { + traversal(root); + return root; + } +} +``` +```java +// 迭代法 +class Solution { + public Node connect(Node root) { + if (root == null) return root; + Queue que = new LinkedList(); + que.offer(root); + Node nodePre = null; + Node node = null; + while (!que.isEmpty()) { + int size = que.size(); + for (int i=0; i 'Node': + def traversal(cur: 'Node') -> 'Node': + if not cur: return [] + if cur.left: cur.left.next = cur.right # 操作1 + if cur.right: + if cur.next: + cur.right.next = cur.next.left # 操作2 + else: + cur.right.next = None + traversal(cur.left) # 左 + traversal(cur.right) # 右 + traversal(root) + return root +``` +```python +# 迭代法 +class Solution: + def connect(self, root: 'Node') -> 'Node': + if not root: return + res = [] + queue = [root] + while queue: + size = len(queue) + for i in range(size): # 开始每一层的遍历 + if i==0: + nodePre = queue.pop(0) # 记录一层的头结点 + node = nodePre + else: + node = queue.pop(0) + nodePre.next = node # 本层前一个节点next指向本节点 + nodePre = nodePre.next + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + nodePre.next = None # 本层最后一个节点指向None + return root ``` - ## Go ```go diff --git a/problems/0714.买卖股票的最佳时机含手续费.md b/problems/0714.买卖股票的最佳时机含手续费.md index 6dc95a2d..b0a64bd2 100644 --- a/problems/0714.买卖股票的最佳时机含手续费.md +++ b/problems/0714.买卖股票的最佳时机含手续费.md @@ -216,7 +216,30 @@ class Solution: # 贪心思路 ``` Go： - +```golang +func maxProfit(prices []int, fee int) int { + var minBuy int = prices[0] //第一天买入 + var res int + for i:=0;i=minBuy&&prices[i]-fee-minBuy<=0{ + continue + } + //可以售卖了 + if prices[i]>minBuy+fee{ + //累加每天的收益 + res+=prices[i]-minBuy-fee + //更新最小值（如果还在收获利润的区间里，表示并不是真正的卖出，而计算利润每次都要减去手续费，所以要让minBuy = prices[i] - fee;，这样在明天收获利润的时候，才不会多减一次手续费！） + minBuy=prices[i]-fee + } + } + return res +} +``` Javascript: ```Javascript // 贪心思路 diff --git a/problems/0738.单调递增的数字.md b/problems/0738.单调递增的数字.md index 7cb24236..82f7f3ae 100644 --- a/problems/0738.单调递增的数字.md +++ b/problems/0738.单调递增的数字.md @@ -159,7 +159,26 @@ class Solution: ``` Go： - +```golang +func monotoneIncreasingDigits(N int) int { + s := strconv.Itoa(N)//将数字转为字符串，方便使用下标 + ss := []byte(s)//将字符串转为byte数组，方便更改。 + n := len(ss) + if n <= 1 { + return N + } + for i:=n-1 ; i>0; i-- { + if ss[i-1] > ss[i] {//前一个大于后一位,前一位减1，后面的全部置为9 + ss[i-1] -= 1 + for j := i ; j < n; j++ {//后面的全部置为9 + ss[j] = '9' + } + } + } + res, _ := strconv.Atoi(string(ss)) + return res +} +``` Javascript: ```Javascript var monotoneIncreasingDigits = function(n) { diff --git a/problems/0925.长按键入.md b/problems/0925.长按键入.md index 4d3543f4..ef712252 100644 --- a/problems/0925.长按键入.md +++ b/problems/0925.长按键入.md @@ -8,7 +8,7 @@

欢迎大家参与本项目，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！

# 925.长按键入 - +题目链接：https://leetcode-cn.com/problems/long-pressed-name/ 你的朋友正在使用键盘输入他的名字 name。偶尔，在键入字符 c 时，按键可能会被长按，而字符可能被输入 1 次或多次。 你将会检查键盘输入的字符 typed。如果它对应的可能是你的朋友的名字（其中一些字符可能被长按），那么就返回 True。 @@ -100,7 +100,31 @@ public: Java： Python： - +```python +class Solution: + def isLongPressedName(self, name: str, typed: str) -> bool: + i, j = 0, 0 + m, n = len(name) , len(typed) + while i< m and j < n: + if name[i] == typed[j]: # 相同时向后匹配 + i += 1 + j += 1 + else: # 不相同 + if j == 0: return False # 如果第一位不相同，直接返回false + # 判断边界为n-1,若为n会越界,例如name:"kikcxmvzi" typed:"kiikcxxmmvvzzz" + while j < n - 1 and typed[j] == typed[j-1]: j += 1 + if name[i] == typed[j]: + i += 1 + j += 1 + else: return False + # 说明name没有匹配完 + if i < m: return False + # 说明type没有匹配完 + while j < n: + if typed[j] == typed[j-1]: j += 1 + else: return False + return True +``` Go： JavaScript： diff --git a/problems/1382.将二叉搜索树变平衡.md b/problems/1382.将二叉搜索树变平衡.md index b9d4fb65..758d5ad8 100644 --- a/problems/1382.将二叉搜索树变平衡.md +++ b/problems/1382.将二叉搜索树变平衡.md @@ -55,7 +55,7 @@ private: vec.push_back(cur->val); traversal(cur->right); } - 有序数组转平衡二叉树 + // 有序数组转平衡二叉树 TreeNode* getTree(vector& nums, int left, int right) { if (left > right) return nullptr; int mid = left + ((right - left) / 2); @@ -76,9 +76,53 @@ public: # 其他语言版本 Java： - +```java +class Solution { + ArrayList res = new ArrayList(); + // 有序树转成有序数组 + private void travesal(TreeNode cur) { + if (cur == null) return; + travesal(cur.left); + res.add(cur.val); + travesal(cur.right); + } + // 有序数组转成平衡二叉树 + private TreeNode getTree(ArrayList nums, int left, int right) { + if (left > right) return null; + int mid = left + (right - left) / 2; + TreeNode root = new TreeNode(nums.get(mid)); + root.left = getTree(nums, left, mid - 1); + root.right = getTree(nums, mid + 1, right); + return root; + } + public TreeNode balanceBST(TreeNode root) { + travesal(root); + return getTree(res, 0, res.size() - 1); + } +} +``` Python： - +```python +class Solution: + def balanceBST(self, root: TreeNode) -> TreeNode: + res = [] + # 有序树转成有序数组 + def traversal(cur: TreeNode): + if not cur: return + traversal(cur.left) + res.append(cur.val) + traversal(cur.right) + # 有序数组转成平衡二叉树 + def getTree(nums: List, left, right): + if left > right: return + mid = left + (right -left) // 2 + root = TreeNode(nums[mid]) + root.left = getTree(nums, left, mid - 1) + root.right = getTree(nums, mid + 1, right) + return root + traversal(root) + return getTree(res, 0, len(res) - 1) +``` Go： JavaScript： diff --git a/problems/二叉树理论基础.md b/problems/二叉树理论基础.md index abeeae67..edd1fed4 100644 --- a/problems/二叉树理论基础.md +++ b/problems/二叉树理论基础.md @@ -223,6 +223,14 @@ type TreeNode struct { } ``` +JavaScript： +``` +function TreeNode(val, left, right) { + this.val = (val===undefined ? 0 : val) + this.left = (left===undefined ? null : left) + this.right = (right===undefined ? null : right) +} +``` -----------------------