feat: update 5 leetcode

Author: kyriejoshua

leetcode/100.same-tree.ts

@@ -0,0 +1,37 @@
+/*
+ * @lc app=leetcode id=100 lang=typescript
+ *
+ * [100] Same Tree
+ */
+
+// @lc code=start
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ * val: number
+ * left: TreeNode | null
+ * right: TreeNode | null
+ * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ * this.val = (val===undefined ? 0 : val)
+ * this.left = (left===undefined ? null : left)
+ * this.right = (right===undefined ? null : right)
+ * }
+ * }
+ */
+
+/**
+ * @description: 深度优先的递归的写法，比较取巧
+ * 核心是找到递归公式和终止条件
+ * 空间和时间复杂度都是： O(min(m, n))
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {
+ if (p === null && q === null) return true;
+ else if (p === null || q === null) return false;
+ else if (p.val !== q.val) return false;
+
+ return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+};
+// @lc code=end

leetcode/1008.construct-binary-search-tree-from-preorder-traversal.ts

@@ -0,0 +1,66 @@
+/*
+ * @lc app=leetcode id=1008 lang=typescript
+ *
+ * [1008] Construct Binary Search Tree from Preorder Traversal
+ */
+
+// @lc code=start
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ * val: number
+ * left: TreeNode | null
+ * right: TreeNode | null
+ * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ * this.val = (val===undefined ? 0 : val)
+ * this.left = (left===undefined ? null : left)
+ * this.right = (right===undefined ? null : right)
+ * }
+ * }
+ */
+
+/**
+ * @description: 递归的思路
+ * @param {number} preorder
+ * @return {TreeNode|null}
+ */
+function bstFromPreorder(preorder: number[]): TreeNode | null {
+ /**
+ * @description: 查找临界点
+ * @param {number} preorder
+ * @param {number} start
+ * @param {number} end
+ * @param {number} val
+ * @return {number}
+ */
+ function findRightIndex(preorder: number[], start: number, end: number, val: number): number {
+ while (start <= end && preorder[start] <= val) {
+ start++;
+ }
+ return start;
+ }
+ /**
+ * @description: 递归构建生成二叉搜索树
+ * @param {number[]} preorder
+ * @param {number} start
+ * @param {number} end
+ * @return {TreeNode | null}
+ */
+ function constructorBST(preorder: number[], start: number, end: number): TreeNode | null {
+ if (start > end) return null;
+ // 取第一个节点作为当前节点构造
+ const root = new TreeNode(preorder[start]);
+ // 说明递归结束
+ if (start === end) return root;
+ // 找到第一个比当前节点的值更大的索引，也就是右子节点的索引
+ const rightIndex = findRightIndex(preorder, start + 1, end, root.val);
+ // 根据这个索引的位置区分当前节点的左右子树的分界点
+ root.left = constructorBST(preorder, start + 1, rightIndex - 1);
+ root.right = constructorBST(preorder, rightIndex, end);
+
+ return root;
+ }
+
+ return constructorBST(preorder, 0, preorder.length - 1);
+};
+// @lc code=end

leetcode/1009.complement-of-base-10-integer.ts

@@ -0,0 +1,44 @@
+/*
+ * @lc app=leetcode id=1009 lang=typescript
+ *
+ * [1009] Complement of Base 10 Integer
+ */
+
+// @lc code=start
+/**
+ * @description: 本题和 476 相同，再做一遍巩固
+ * 这是按位非的实现
+ * 时间复杂度 O(logn)
+ * 空间复杂度 O(1)
+ * @param {number} n
+ * @return {number}
+ */
+function bitwiseComplement(n: number): number {
+ let highbit = 0;
+
+ // 从 1 开始循环
+ for (let i = 1; i <= 30; i++) {
+ // 通过循环找到最高的位 i; 左移就是放大数字来查找最高位
+ if (n >= 1 << i) {
+ highbit = i;
+ } else {
+ break;
+ }
+ }
+ // 防止溢出
+ const mask = highbit === 30 ? 0x7fffffff : (1 << (highbit + 1)) - 1;
+
+ return n ^ mask;
+};
+
+function bitwiseComplement2(n: number): number {
+ const str = n.toString(2); // 转成 2 进制
+ let reversed = '';
+
+ for (const s of str) {
+ reversed += s === '1' ? '0' : '1';
+ }
+
+ return parseInt(reversed, 2);
+};
+// @lc code=end

leetcode/557.reverse-words-in-a-string-iii.ts

@@ -0,0 +1,20 @@
+/*
+ * @lc app=leetcode id=557 lang=typescript
+ *
+ * [557] Reverse Words in a String III
+ */
+
+// @lc code=start
+/**
+ * @description: 直接使用数组 API
+ * @param {string} s
+ * @return {string}
+ */
+function reverseWords(s: string): string {
+ let str = '';
+ for (const word of s.split(' ')) {
+ str += word.split('').reverse().join('') + ' ';
+ }
+ return str.trim();
+};
+// @lc code=end

leetcode/617.merge-two-binary-trees.ts

@@ -0,0 +1,37 @@
+/*
+ * @lc app=leetcode id=617 lang=typescript
+ *
+ * [617] Merge Two Binary Trees
+ */
+
+// @lc code=start
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ * val: number
+ * left: TreeNode | null
+ * right: TreeNode | null
+ * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ * this.val = (val===undefined ? 0 : val)
+ * this.left = (left===undefined ? null : left)
+ * this.right = (right===undefined ? null : right)
+ * }
+ * }
+ */
+
+/**
+ * @description: 递归实现，核心在于确认递推公式
+ * 时间和空间复杂度一样，都是 O(min(m, n))
+ * @param {TreeNode|null} root1
+ * @param {TreeNode|null} root2
+ * @return {TreeNode|null}
+ */
+function mergeTrees(root1: TreeNode | null, root2: TreeNode | null): TreeNode | null {
+ if (root1 === null) return root2;
+ if (root2 === null) return root1;
+ const node: TreeNode | null = new TreeNode(root1.val + root2.val);
+ node.left = mergeTrees(root1.left, root2.left);
+ node.right = mergeTrees(root1.right, root2.right);
+ return node;
+};
+// @lc code=end