3-无重复字符的最长子串(Longest Substring Without Repeating Characters)

原文链接: https://leetcode-cn.com/problems/longest-substring-without-repeating-characters

英文原文

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

Example 1:

Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Example 4:

Input: s = ""
Output: 0

Constraints:

0 <= s.length <= 5 * 10⁴
s consists of English letters, digits, symbols and spaces.

中文题目

给定一个字符串 s ，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: s = "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: s = "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: s = "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。

示例 4:

输入: s = ""
输出: 0

提示：

0 <= s.length <= 5 * 10⁴
s 由英文字母、数字、符号和空格组成

通过代码

高赞题解

思路：

这道题主要用到思路是：滑动窗口

什么是滑动窗口？

其实就是一个队列,比如例题中的 abcabcbb，进入这个队列（窗口）为 abc 满足题目要求，当再进入 a，队列变成了 abca，这时候不满足要求。所以，我们要移动这个队列！

如何移动？

我们只要把队列的左边的元素移出就行了，直到满足题目要求！

一直维持这样的队列，找出队列出现最长的长度时候，求出解！

时间复杂度：$O(n)$

代码：

[1]
class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        if not s:return 0
        left = 0
        lookup = set()
        n = len(s)
        max_len = 0
        cur_len = 0
        for i in range(n):
            cur_len += 1
            while s[i] in lookup:
                lookup.remove(s[left])
                left += 1
                cur_len -= 1
            if cur_len > max_len:max_len = cur_len
            lookup.add(s[i])
        return max_len

[1]
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        if(s.size() == 0) return 0;
        unordered_set<char> lookup;
        int maxStr = 0;
        int left = 0;
        for(int i = 0; i < s.size(); i++){
            while (lookup.find(s[i]) != lookup.end()){
                lookup.erase(s[left]);
                left ++;
            }
            maxStr = max(maxStr,i-left+1);
            lookup.insert(s[i]);
    }
        return maxStr;
        
    }
};

[1]
class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s.length()==0) return 0;
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        int max = 0;
        int left = 0;
        for(int i = 0; i < s.length(); i ++){
            if(map.containsKey(s.charAt(i))){
                left = Math.max(left,map.get(s.charAt(i)) + 1);
            }
            map.put(s.charAt(i),i);
            max = Math.max(max,i-left+1);
        }
        return max;
        
    }
}

下面介绍关于滑动窗口的万能模板,可以解决相关问题,相信一定可以对滑动窗口有一定了解!

模板虽好,还是少套为好!多思考!更重要!

还有类似题目有:

3. 无重复字符的最长子串

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        from collections import defaultdict
        lookup = defaultdict(int)
        start = 0
        end = 0
        max_len = 0
        counter = 0
        while end < len(s):
            if lookup[s[end]] > 0:
                counter += 1
            lookup[s[end]] += 1
            end += 1
            while counter > 0:
                if lookup[s[start]] > 1:
                    counter -= 1
                lookup[s[start]] -= 1
                start += 1
            max_len = max(max_len, end - start)
        return max_len

76. 最小覆盖子串

class Solution:
    def minWindow(self, s: 'str', t: 'str') -> 'str':
        from collections import defaultdict
        lookup = defaultdict(int)
        for c in t:
            lookup[c] += 1
        start = 0
        end = 0
        min_len = float("inf")
        counter = len(t)
        res = ""
        while end < len(s):
            if lookup[s[end]] > 0:
                counter -= 1
            lookup[s[end]] -= 1
            end += 1
            while counter == 0:
                if min_len > end - start:
                    min_len = end - start
                    res = s[start:end]
                if lookup[s[start]] == 0:
                    counter += 1
                lookup[s[start]] += 1
                start += 1
        return res

159. 至多包含两个不同字符的最长子串

class Solution:
    def lengthOfLongestSubstringTwoDistinct(self, s: str) -> int:
        from collections import defaultdict
        lookup = defaultdict(int)
        start = 0
        end = 0
        max_len = 0
        counter = 0
        while end < len(s):
            if lookup[s[end]] == 0:
                counter += 1
            lookup[s[end]] += 1
            end +=1
            while counter > 2:
                if lookup[s[start]] == 1:
                    counter -= 1
                lookup[s[start]] -= 1
                start += 1
            max_len = max(max_len, end - start)
        return max_len

340. 至多包含 K 个不同字符的最长子串

class Solution:
    def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
        from collections import defaultdict
        lookup = defaultdict(int)
        start = 0
        end = 0
        max_len = 0
        counter = 0
        while end < len(s):
            if lookup[s[end]] == 0:
                counter += 1
            lookup[s[end]] += 1
            end += 1
            while counter > k:
                if lookup[s[start]] == 1:
                    counter -= 1
                lookup[s[start]] -= 1
                start += 1
            max_len = max(max_len, end - start)
        return max_len