Longest Substring Without Repeating Characters

最新推荐文章于 2025-08-07 20:58:17 发布

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本文探讨了如何寻找字符串中最长的无重复字符子串，提供了两种解决方案：暴力破解法和滑动窗口法。通过实例解析，展示了不同方法的时间复杂度及优劣对比。

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Longest Substring Without Repeating Characters

Description

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

Example

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.

Solution

第一种方法,暴力破解法

public class Solution {
    public static int lengthOfLongestSubString(String s){
        int n = s.length();
        int ans = 0;
        // 从第一个字符开始遍历,发现有重复的就返回
        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                if (unique(s,i,j)){
                    ans = Math.max(ans,j-i);
                }
            }
        }
        return ans;
    }
    public static boolean unique(String s,int i,int j){
        HashSet<Character> ss = new HashSet<>();
        for (int k = i; k < j; k++) {
            Character c = s.charAt(k);
            if (ss.contains(c)){
                return false;
            }
            ss.add(c);
        }
        return true;
    }
    public static void main(String[] args) {
        String s1 = "abcabcbb";
        String s2 = "bbbbb";
        String s3 = "pwwkew";
        System.out.println(lengthOfLongestSubString(s3));
    }
}

时间复杂度O(n^2)

滑动窗口法

public static int lengthOfLongestSubString(String s) {
     int n = s.length();
     int ans = 0;
     // 当前index
     HashMap<Character, Integer> map = new HashMap<>();
     // [i,j]窗口在这之间滑动
     for (int i = 0, j = 0; j < n; j++) {
         if (map.containsKey(s.charAt(j))){
             i = Math.max(map.get(s.charAt(j)),i);
         }
         ans = Math.max(ans,j-i+1);
         map.put(s.charAt(j),j+1);
     }
     return ans;
 }