LeetCode：3.无重复字符的最长子串

LeetCode：3.无重复字符的最长子串

优化用kmp

解题步骤用双指针维护一个滑动窗囗，用来剪切子串。不断移动右指针，遇到重复字符，就把左指针移动到重复字符的下一位。过程中，记录所有窗口的长度，并返回最大值。

时间复杂度：O(n)空间复杂度：O(m)，m是字符串中不重复字符的个数

var lengthOfLongestSubstring = function(s) {let str = '';let longestStr = '';for (let i = 0; i < s.length; i++) {if (!str.includes(s[i])) {str += s[i];} else {// Check if the current substring is longer than the longest found so farif (str.length > longestStr.length) {longestStr = str;}// Find the position of the repeated character in the current substringconst repeatIndex = str.indexOf(s[i]);// Start a new substring from the next character after the repeated onestr = str.substring(repeatIndex + 1) + s[i];}}// Final check to update the longest substring if neededif (str.length > longestStr.length) {longestStr = str;}console.log(longestStr);return longestStr.length;
};
console.log(lengthOfLongestSubstring('pwfwkew'))
//只能用 string 和数组  time O(n*n)  space O(n)//Map Solution time O(n)  space O(m)
var lengthOfLongestSubstring = function(s) {let oneNeedle=0let twoNeedle=0let map=new Map()for(let i=0;i<s.length;i++){if(map.has(s[i])&&map.get(s[i])>=oneNeedle){oneNeedle=map.get(s[i])+1}twoNeedle=Math.max(twoNeedle,i-oneNeedle+1)map.set(s[i],i)}return twoNeedle
};
console.log(lengthOfLongestSubstring('pwfwkew'))

flowchart TDA[开始] --> B{是否遍历结束}B -->|No| C[获取当前字符]C --> D{字符是否存在于 map 且位置 >= oneNeedle}D -->|Yes| E[更新 oneNeedle]D -->|No| F[跳过更新 oneNeedle]E --> G[更新最大长度]F --> GG --> H[更新 map]H --> BB -->|Yes| I[返回最大长度]

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