LeetCode——Longest Substring Without Repeating Characters

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: 1
Explanation: The answer is "b", with the length of 1.

Example 3:
Input: "pwwkew"
Output: 3
Explanation: 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...