Merge branch 'gh-pages' of https://github.com/guocaoyi/leetcode-cn

Author: guocaoyi

Algorithms/0001.Two Sum/index.js

@@ -0,0 +1,63 @@
+/**
+ * 方案一：两次列表遍历
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number[]}
+ */
+export function onTwiceBubble(
+ nums: Array<number>,
+ target: number
+): Array<number> {
+ for (let i = 0; i < nums.length; i++) {
+ for (let j = i + 1; j < nums.length; j++) {
+ if (nums[i] + nums[j] == target) {
+ return [i, j];
+ }
+ }
+ }
+}
+
+/**
+ * 方案二：两次哈希遍历
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number[]}
+ */
+export function onTwiceIteratorHash(
+ nums: Array<number>,
+ target: number
+): Array<number> {
+ const map: any = {};
+ const length = nums.length;
+ // 使用Array.forEach在性能上会有点损耗(测试用例:61ms到59ms)
+ for (let i = 0; i < length; i++) {
+ map[nums[i]] = i;
+ }
+ for (let i = 0; i < length; i++) {
+ const x = target - nums[i];
+ if (x in map && map[x] != i) {
+ return [i, map[x]];
+ }
+ }
+}
+
+/**
+ * 方案三：一次哈希遍历
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number[]}
+ */
+export function onIteratorHash(
+ nums: Array<number>,
+ target: number
+): Array<number> {
+ const map: any = {};
+ const length = nums.length;
+ for (let i = 0; i < length; i++) {
+ const x = target - nums[i];
+ if (x in map && map[x] != i) {
+ return [map[x], i];
+ }
+ map[nums[i]] = i;
+ }
+}

Algorithms/0001.Two Sum/index.ts

@@ -0,0 +1,44 @@
+/**
+ * @param {string} s
+ * @return {number}
+ * 987test case,132ms
+ */
+const slution1 = (s: string): number => {
+ let maxSub = "",
+ currentSub = "";
+ const arr = s.split("");
+ arr.forEach((s: string) => {
+ if (currentSub.includes(s)) {
+ // 存在
+ if (currentSub.length >= maxSub.length) {
+ maxSub = currentSub;
+ }
+ let [lStr, rStr] = currentSub.split(s);
+ currentSub = rStr || "";
+ currentSub += s;
+ } else {
+ // 不存在
+ currentSub += s;
+ if (currentSub.length >= maxSub.length) {
+ maxSub = currentSub;
+ }
+ }
+ });
+ return maxSub.length;
+};
+
+/**
+ * 987test case,184ms
+ */
+let slution2 = (s: string): number => {
+ const obj = {
+ subStr: "",
+ maxLen: 0
+ };
+ for (let i: number = 0; i < s.length; i++) {
+ let strArray: string[] = obj.subStr.split(s[i]);
+ obj.subStr = strArray[strArray.length - 1] + s[i];
+ obj.maxLen = Math.max(obj.maxLen, obj.subStr.length);
+ }
+ return obj.maxLen;
+};

Algorithms/0003.Longest Substring Without Repeating Characters/README.md

@@ -1,11 +1,8 @@
-"use strict"
+# 将字符串 "PAYPALISHIRING" 以 Z 字形排列成给定的行数：
 
-/**
- * 将字符串 "PAYPALISHIRING" 以Z字形排列成给定的行数：
-
-P A H N
+P A H N
 A P L S I I G
-Y  I  R
+Y I R
 之后从左往右，逐行读取字符："PAHNAPLSIIGYIR"
 
 实现一个将字符串进行指定行数变换的函数:
@@ -21,10 +18,7 @@ string convert(string s, int numRows);
 输出: "PINALSIGYAHRPI"
 解释:
 
-P I N
-A L S I G
-Y A H R
-P I
-*/
-
-
+P I N
+A L S I G
+Y A H R
+P I

Algorithms/0003.Longest Substring Without Repeating Characters/index.ts

@@ -0,0 +1,3 @@
+/**
+ *
+ */

Algorithms/0003.Longest Substring Without Repeating Characters/longestSubstringWithoutRepeatingCharacters.js

@@ -94,7 +94,7 @@ var longestCommonPrefix = function(strs) {
 
 ### III: 单次遍历
 
-使用`strs[0]`作为初始前缀串，逐一遍历`strs[]`元素进行比较，如`String.indexOf === -1`则自减长度 1，直至为 0 成立后继续访问下面的元素。
+使用 `strs[0]` 作为初始前缀串，逐一遍历 `strs[]` 元素进行比较，如 `String.indexOf !== 0` 则自减长度 1，直至成立后继续访问后面的元素。
 
 - language: javascript
 - status: Accepted