Longest Substring no Repeat Characters-德赢Vwin官网网

Longest Substring no Repeat Characters

今天的题目是第3题,medium难度.

Longest Substring Without Repeating Charac te rs

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.

3. 无重复字符的最长子串

给定一个字符串，请你找出其中不含有重复字符的最长子串的长度。

示例 1:

输入: "abcabcbb"
输出: 3 
解释: 因为无重复字符的最长子串是 "abc", 所以其长度为3。

示例 2:

输入: "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b"，所以其长度为1。

示例 3:

输入: "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke"，所以其长度为3。请注意，你的答案必须是 子串 的长度，"pwke"是一个子序列，不是子串。

My answer:

首先对于查询是否存在的操作我们选择用dict来做(hash速度快), 对整个字符串进行遍历用dict字典中存储已经访问过的数据. 对于未存在于dict中的元素直接添加key:value为s[i]:i; 当遇到已经存在的元素更新start的位置为dict[s[i]]的下一位, 因为dict中的值仍然保留start之前的数元素, 所以遇到的存在元素未必是有效的, 需要对start的更新值进行判断start = max(start, dct[s[i]] + 1). 最后更字典和最大长度即可.

class Solution:


    def lengthOfLongestSubstring(self, s: str) -> int:
        start = 0
        max_len = 0
        dct = {}
        for i in range(len(s)):
            if s[i] in dct:
                start = max(start, dct[s[i]] + 1) 
            dct[s[i]] = i
            max_len = max(max_len, i - start+1)
        return max_len