一、题目说明

二、求解方法

2.1 暴力法

1、思路

2、代码

package question4.solution1;

import java.util.HashSet;

/**
 * @description: 暴力法
 * @author: hyr
 * @time: 2020/3/26 15:40
 */


public class Solution {
    public static void main(String[] args) {
        String s = "abcccccaaaa";
        System.out.println(lengthOfLongestSubstring(s));
    }

    public static int lengthOfLongestSubstring(String s){
        int n = s.length();
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1 ; j <= n ; j++) {
                if (allUnique(s, i, j)){
                    // 这里是比较这一次与上一次的子串长度
                    ans = Math.max(ans, j - i);
                }
            }
        }
        return ans;
    }

    public static boolean allUnique(String s, int start, int end){
        HashSet<Character> set = new HashSet<>();
        for (int i = start; i < end; i++) {
            char ch = s.charAt(i);
            if (set.contains(ch)){
                return false;
            }
            set.add(ch);
        }
        return true;
    }
}

2.2 滑动窗口法

1、思路

2、代码

package question4.solution2;

import java.util.HashSet;
import java.util.Set;

/**
 * @description: 滑动窗口法
 * @author: hyr
 * @time: 2020/3/26 15:40
 */
public class Solution {
    public static void main(String[] args) {
        String s = "abccccaabcdeaa";
        System.out.println(lengthOfLongestSubstring(s));
    }

    public static int lengthOfLongestSubstring(String s) {
        int n = s.length();
        Set<Character> set = new HashSet<>();
        int ans = 0, i = 0, j = 0;
        while (i < n && j < n) {
            // try to extend the range [i, j]
            if (!set.contains(s.charAt(j))) {
                set.add(s.charAt(j++));
                ans = Math.max(ans, j - i);
            } else {
                set.remove(s.charAt(i++));
            }
        }
        return ans;
    }
}