LeetCode-3：Longest Substring Without Repeating Characters

一、题目

原题链接：https://leetcode.com/problems/longest-substring-without-repeating-characters/
难度等级：中等

Given a string, find the length of the longest substring without repeating characters.

Examples:
Given “abcabcbb”, the answer is “abc”, which the length is 3.
Given “bbbbb”, the answer is “b”, with the length of 1.

Given “pwwkew”, 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.

题目翻译：
给定一个字符串，找出其中无重复字符的最长子串的长度。
例如：
“abcabcbb”，无重复字符的最长子串是”abc”，长度为3。
“bbbbb”，无重复字符的最长子串是”b”，长度为1

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二、解题方案

思路1：
通过Bitmap来统计是否子字符串中有重复的字符。遍历时若遇到重复值，需要回溯。

代码实现：

class Sulution {
public:
    int lengthOfLongestSubstring(string s) {
        map<char, int> m;
        int maxLen = 0;
        int lastRepeatPos = -1;
        for (int i = 0; i < s.size(); ++i) {
            if (m.find(s[i]) != m.end() && lastRepeatPos < m[s[i]]) {
                lastRepeatPos = m[s[i]];
            }
            if (i - lastRepeatPos > maxLen) {
                maxLen = i -lastRepeatPos;
            }
            m[s[i]] = i;
        }

        return maxLen;
    }
};

思路2：
通过一个数字来存储每一个字符上一次出现的位置，当出现冲突时及时调整。

代码实现：

class Sulution {
public:
    int lengthOfLongestSubstring(string s) {
        const int MAX_CHARS = 256;
        int m[MAX_CHARS];
        memset(m, -1, sizeof(m));

        int maxLen = 0;
        int lastRepeatPos = -1;
        for (int i = 0; i < s.size(); ++i) {
            if (m[s[i]] != -1 && lastRepeatPos < m[s[i]]) {
                lastRepeatPos = m[s[i]];
            }
            if (i - lastRepeatPos > maxLen) {
                maxLen = i -lastRepeatPos;
            }
            m[s[i]] = i;
        }

        return maxLen;
    }
};