LeetCode-3：Longest Substring Without Repeating Characters (最长无重复字符的子串) --medium_longest substring without repeating characters pyt-优快云博客

本文链接：https://blog.youkuaiyun.com/koala_tree/article/details/78749798

本文介绍了一种使用动态规划方法解决寻找字符串中最长无重复字符子串的问题，提供了C++和Python两种语言的实现方案。

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题目：

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

Example:

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.

问题解析：

给定一个字符串，从中找出不含重复字符的最长子串的长度。

链接：

https://leetcode.com/problems/longest-substring-without-repeating-characters/description/

思路标签：

算法：DP

数据结构：Vector

解答：

1. C++

假设L[i] = s[start … i]，表示最长的子字符串，其中没有重复的元素，我们以Vector作为hashmap保存< 字符，索引>；
然后访问s[i + 1]：
- 1）如果s[i + 1]没有出现在散列表dict中，我们可以将s[i + 1]添加到散列表中，则有L[i + 1] = s[start … i + 1]；
- 2）如果s[i + 1]存在于dict中，并且dict中对应的索引是k；则令start = max(start，k)，则L[i + 1] = s[start … i + 1]，记录此时的最大长度；
整个问题就是围绕不加入重复元素而展开的。
时间复杂度：O(n)

class Solution {
public:
    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;
    }
};

2. python

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        start = maxLength = 0
        usedChar = {}

        for i in range(len(s)):
            if s[i] in usedChar and start <= usedChar[s[i]]:
                start = usedChar[s[i]] + 1
            else:
                maxLength = max(maxLength, i - start + 1)

            usedChar[s[i]] = i

        return maxLength