LeetCode 3. Longest Substring Without Repeating-优快云博客

本文介绍了一种算法，用于解决字符串中无重复字符最长子串的问题，通过使用哈希映射和双指针技巧，实现高效求解。

题目：

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

思路：

用一个map存放字符出现的position；用两个指针begin,end指向substr的首末；遇到相同字符时将begin移到该字符之前出现的后一位置，并更新该字符的position。

代码：C++ :

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if (s.length() <= 1)
            return s.length();
        
        int begin = 0;
        int end = 0;
        map<char,int> mapping;
        int solve = 0;
        
        mapping[s[0]] = 0;
        while (end < s.length() - 1) {
            end++;
           if (mapping.find(s[end]) != mapping.end()) {
                int gap = end - begin;
                if (gap > solve)
                    solve = gap;
                begin = mapping[s[end]] + 1 > begin ? mapping[s[end]] + 1 : begin;
                mapping[s[end]] = end;
            }
            else
                mapping[s[end]] = end;
        }
        int gap = end - begin + 1;
        if (gap > solve)
            solve = gap;
        return solve;
    }
};

Reference : https://leetcode.com/discuss/59051/c-code-in-9-lines

本质上跟我的思路差不多，但是代码很简洁

int lengthOfLongestSubstring(string s) {
        vector<int> dict(256, -1);
        int maxLen = 0, start = -1;
        for (int i = 0; i != s.length(); i++) {
            if (dict[s[i]] > start)
                start = dict[s[i]];
            dict[s[i]] = i;
            maxLen = max(maxLen, i - start);
        }
        return maxLen;
    }