update 7 leetcode

Author: kyriejoshua

leetcode/145.binary-tree-postorder-traversal.js

@@ -0,0 +1,70 @@
+/*
+ * @lc app=leetcode id=145 lang=javascript
+ *
+ * [145] Binary Tree Postorder Traversal
+ */
+
+// @lc code=start
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ * this.val = (val===undefined ? 0 : val)
+ * this.left = (left===undefined ? null : left)
+ * this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * 迭代循环的方式
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+var postorderTraversal = function(root) {
+ if (!root) {
+ return [];
+ }
+ let queues = [];
+ let res = [];
+ let prev = null;
+
+ while (queues.length || root) {
+ // 把左侧的全部推入
+ while (root !== null) {
+ queues.push(root);
+ root = root.left;
+ }
+
+ // 后序遍历，一定用到的是栈的思路
+ root = queues.pop();
+ // 遍历到末端了
+ if (root.right == null || root.right == prev) {
+ res.push(root.val);
+ prev = root;
+ root = null;
+ } else {
+ // 否则则继续添加入右子树
+ queues.push(root);
+ root = root.right;
+ }
+ }
+ return res;
+};
+
+/**
+ * 递归的方式
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+var postorderTraversal2 = function(root) {
+ let res = [];
+ function walkTree(root) {
+ if (!root) {
+ return;
+ }
+ root.left && walkTree(root.left);
+ root.right && walkTree(root.right);
+ res.push(root.val);
+ }
+ walkTree(root);
+ return res;
+};
+// @lc code=end

leetcode/1486.xor-operation-in-an-array.js

@@ -0,0 +1,24 @@
+/*
+ * @lc app=leetcode id=1486 lang=javascript
+ *
+ * [1486] XOR Operation in an Array
+ */
+
+// @lc code=start
+/**
+ * @param {number} n
+ * @param {number} start
+ * @return {number}
+ */
+var xorOperation = function(n, start) {
+ let i = 0;
+ let sum = start;
+ while (i <= n - 1) {
+ if (i !== 0) {
+ sum = sum ^ (start + 2 * i);
+ }
+ i++;
+ }
+ return sum;
+};
+// @lc code=end

leetcode/202.happy-number.js

@@ -0,0 +1,36 @@
+/*
+ * @lc app=leetcode id=202 lang=javascript
+ *
+ * [202] Happy Number
+ */
+
+// @lc code=start
+/**
+ * 双指针，龟兔赛跑
+ * @param {number} n
+ * @return {boolean}
+ */
+var isHappy = function(n) {
+ // 根据规律分析可得，最终的值要么会进入一个循环，要么最终会返回 1
+ function goNext(n) {
+ let sum = 0;
+ while (n > 0) {
+ const d = n % 10;
+ n = Math.floor(n / 10);
+ sum += d * d;
+ }
+ return sum;
+ }
+ let slow = n;
+ let fast = goNext(n);
+ // 快慢指针查看是否进入循环
+ while (slow !== fast && fast !== 1) {
+ console.info('slow', slow);
+ console.info('fast', fast);
+ slow = goNext(slow);
+ fast = goNext(goNext(fast));
+ }
+ return fast === 1;
+};
+isHappy(19);
+// @lc code=end

leetcode/283.move-zeroes.js

@@ -0,0 +1,54 @@
+/*
+ * @lc app=leetcode id=283 lang=javascript
+ *
+ * [283] Move Zeroes
+ */
+
+// @lc code=start
+/**
+ * 双指针的思路
+ * @param {number[]} nums
+ * @return {void} Do not return anything, modify nums in-place instead.
+ */
+var moveZeroes = function(nums) {
+ let left = right = 0;
+ let len = nums.length;
+
+ while (right < len) {
+ // 左指针仅在为 0 的情况下，与右边的数交换 ❌ 错的
+ // 右指针在不为 0 的情况下，和左指针交换，这样保证了数都是相对的顺序没有变，而且在转换过程中， 0 自然去了右边
+ if (nums[right] !== 0) {
+ [nums[right], nums[left]] = [nums[left], nums[right]];
+ left++;
+ }
+ // 右指针始终在遍历
+ right++;
+ }
+ return nums;
+};
+
+var moveZeroes2 = function(nums) {
+ let i = 0;
+ let sum = 0;
+ let c = 0;
+
+ // 保存 0 的数量用来当终止条件
+ nums.forEach((s) => {
+ if (s === 0) {
+ sum++;
+ }
+ });
+
+ // 是 0 就推到数组末尾去
+ while (c < sum) {
+ if (nums[i] === 0) {
+ let p = nums.splice(i, 1);
+ nums.push(p);
+ c++;
+ } else {
+ i++;
+ }
+ }
+ return nums;
+};
+// @lc code=end

leetcode/821.shortest-distance-to-a-character.js

@@ -0,0 +1,39 @@
+/*
+ * @lc app=leetcode id=821 lang=javascript
+ *
+ * [821] Shortest Distance to a Character
+ */
+
+// @lc code=start
+/**
+ * @param {string} s
+ * @param {character} c
+ * @return {number[]}
+ */
+var shortestToChar = function(s, c) {
+ let prev = s.length - 1;
+ let res = [];
+ let left = 0;
+
+ // 从左往右遍历，保存一份结果
+ while (left < s.length) {
+ if (s.charAt(left) === c) {
+ prev = left;
+ }
+ res[left] = Math.abs(left - prev);
+ left++;
+ }
+
+ // 再从右往左遍历，保存另一份结果并比较
+ let right = s.length - 1;
+ while (right >= 0) {
+ if (s.charAt(right) === c) {
+ prev = right;
+ }
+
+ res[right] = Math.min(Math.abs(right - prev), res[right]);
+ right--;
+ }
+ return res;
+};
+// @lc code=end

leetcode/844.backspace-string-compare.js

@@ -0,0 +1,108 @@
+/*
+ * @lc app=leetcode id=844 lang=javascript
+ *
+ * [844] Backspace String Compare
+ */
+
+// @lc code=start
+/**
+ * 暴力法，直接循环得出退格后的字符串再比较
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var backspaceCompare = function(s, t) {
+ let left = s.length - 1;
+ let right = t.length - 1;
+ let skipS = 0;
+ let skipT = 0;
+
+ while (left >= 0 || right >= 0) {
+ // 通过算法取出当前需要比较的字符串索引
+ while (left >= 0) {
+ if (s.charAt(left) === '#') {
+ skipS++;
+ left--;
+ } else if (skipS > 0) {
+ skipS--;
+ left--;
+ } else {
+ break;
+ }
+ }
+
+ while (right >= 0) {
+ if (t.charAt(right) === '#') {
+ skipT++;
+ right--;
+ } else if (skipT > 0) {
+ skipT--;
+ right--;
+ } else {
+ break;
+ }
+ }
+
+ // 直接比较对应的字符串是否相等
+ if (left >= 0 && right >= 0) {
+ if (s[left] !== t[right]) {
+ return false;
+ }
+ } else {
+ // 进入这个条件则说明两边长度不一致
+ if (left >= 0 || right >= 0) {
+ return false;
+ }
+ }
+
+ left--;
+ right--;
+ }
+
+ return true;
+};
+
+/**
+ * 暴力法，直接循环得出退格后的字符串再比较
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var backspaceCompare2 = function(s, t) {
+ let left = s.length - 1;
+ let right = t.length - 1;
+ let leftVal = '';
+ let rightVal = '';
+ let backNum = 0;
+
+ while (left >= 0) {
+ if (s[left] === '#') {
+ backNum++;
+ } else {
+ if (backNum) {
+ backNum--;
+ } else {
+ leftVal = `${s[left]}${leftVal}`;
+ }
+ }
+
+ left--;
+ }
+ backNum = 0;
+ while (right >= 0) {
+ if (t[right] === '#') {
+ backNum++;
+ } else {
+ if (backNum) {
+ backNum--;
+ } else {
+ rightVal = `${t[right]}${rightVal}`;
+ }
+ }
+
+ right--;
+ }
+ console.info(leftVal, rightVal);
+ return leftVal === rightVal;
+};
+// @lc code=end