原问题Given a string, find the length of the longest substring without repeating characters.Example 1:Input: “abcabcbb"Output: 3 Explanation: The answer is “abc”, with the length of 3. Example 2:Input: “bbbbb"Output: 1Explanation: The answer is “b”, with the length of 1.Example 3:Input: “pwwkew"Output: 3Explanation: The answer is “wke”, with the length of 3. Note that the answer must be a substring, “pwke” is a subsequence and not a substring.我的沙雕解法var lengthOfLongestSubstring = function(s) { let recordObj = {}; let length = s.length; let max = 0; for(let i = 0; i < length; i++){ let record = 0; for(let g = i; g < length; g++){ if(recordObj[s[g]] === undefined){//无重复字母 recordObj[s[g]] = true; record++; }else{//存在重复字母 recordObj = {}; break; } } max = record > max? record:max; if(max === length){break;} } return max;};挨打：最暴力的无脑解法，耗时672ms。。。贪心解法学习参考了排名较前的答案，多数是贪心思想，以下摘抄一位的代码并加上学习的注释/*** 通过i，j指针计算子序列长度* j指针：表示当前循环的字母，i指针：表示起点* map用于记录出现过的字母的相邻下标，给予i新的起点* 重复字母出现时，比较当前起点与map的记录位置，取较大值，保证连续子序列，同时体现贪心：求* 当前可求得的最长子序列**/var lengthOfLongestSubstring = function(s) { var n = s.length, ans = 0; var map = new Map(); // 记录出现过字母的相邻下标 // try to extend the range [i, j] for (var j = 0, i = 0; j < n; j++) { if (map.has(s.charAt(j))) { //若此字母在之前的循环中出现过 i = Math.max(map.get(s.charAt(j)), i); //保证连续子序列，同时体现贪心 } ans = Math.max(ans, j - i + 1); //比较 map.set(s.charAt(j), j + 1); //记录字母的相邻位置 } return ans;};此算法耗时108ms百度到一张图片很有利于理解 举例：qpxrjxkltzyx图片来源：https://www.cnblogs.com/wangk…