Leetcode 3. Longest Substring Without Repeating Characters思路解析

最长无重复字符子串

最新推荐文章于 2024-01-06 13:50:14 发布

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最新推荐文章于 2024-01-06 13:50:14 发布

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分类专栏： LeetCode 文章标签： leetcode 动态规划 dp

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本文介绍了一种求解给定字符串中最大长度无重复字符子串的方法，并提供了Python实现。通过实例演示了如何迭代更新最大子串及其长度。

题目

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.

思路

先以下面这case举例：
Given “wobgrovw” the answer is “bgrovw”，with the length of 6

令sub_str为当前迭代的最大连续子字符串， sub_set是历史的最大连续子字符串
sub_str sub_set
“w” []
“wo” []
“wob” []
“wobg” []
“wobgr” []
“bgro” [“wobgr”]
“bgrov” [“wobgr”]
“bgrovw” [“wobgr”]
“” [“wobgr”, “bgrovw”]

然后遍历sub_set 中最长的sub_str，即得到答案

Python实现

class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        sub_set = []
        sub_str = ''
        for i in s:
            if i not in sub_str:
                sub_str += i
            elif i == sub_str[0]:
                sub_str = sub_str[1:] + i
            elif i == sub_str[-1]:
                sub_set.append(sub_str)
                sub_str = i
            else:
                sub_set.append(sub_str)
                idx = sub_str.index(i)
                sub_str = sub_str[idx+1:] + i
        if sub_str:
            sub_set.append(sub_str)
        max_sub_len = 0
        for sub_str in sub_set:
            if len(sub_str) > max_sub_len:
                max_sub_len = len(sub_str)
        return max_sub_len