leetcode-3. 无重复字符的最长子串

题目

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。

解题思路

Queue

很简单的一道题目，队列即可。用队列记录当前的子串，如果新字符在队列中，则元素依次出队，直到重复的元素出队为止。

改进版可以考虑使用双端队列，popleft()的时间复杂度同样是$o(1)$

时间复杂度：$o(s)$
空间复杂度：$o(s)$

Sliding Window

Use sliding window could reduce the space complexity from $o(n)$ to $o(1)$, the key is use two pointer to denote the queue above and use a map to keep track of every element we have visited.

Hash Map

Use a hash map to keep track of the character and the index, if it’s duplicate, then move the left to the right of the character that is duplicate with right. Note that both left and right only move forward.

Time complexity: $o(n)$
Space complexity: $o(n)$

代码

Queue

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        queue = []
        max_length = 0
        for ch in s:
            while ch in queue:
                queue.pop(0)
            queue.append(ch)
            max_length = max(max_length, len(queue))
        return max_length

Sliding Window

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        left, right = 0, 0
        res = 0
        visited_char = {}
        for right, ch in enumerate(s):
            while ch in visited_char:
                visited_char.pop(s[left])
                left += 1
            visited_char[ch] = right
            res = max(res, right - left + 1)
        return res

Hash Map

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        char_index = {}
        left = 0
        res = 0
        for right in range(len(s)):
            if s[right] in char_index:
                left = max(left, char_index[s[right]] + 1)
            char_index[s[right]] = right
            res = max(res, right - left + 1)
        return res