LeetCode-Longest Substring Without Repeating Characters(最长不重复子串)

原题:
在这里插入图片描述
思路:

  1. 列举出来所有的合理的子串,初始的索引设为i,结束的索引设为j,设置双层循环,i:0到n-1;j:i+1到n;
  2. 设置一个set,将每次检查后不重复的字符add进去,如果重复的字符,就返回false。

代码一:

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

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();//n代表字符串的长度
        int ans = 0;
        for (int i = 0; i < n; i++)//双层循环嵌套
            for (int j = i + 1; j <= n; j++)
                if (allUnique(s, i, j)) 
                	ans = Math.max(ans, j - i);
        return ans;
    }

    public boolean allUnique(String s, int start, int end) {
        Set<Character> set = new HashSet<>();
        for (int i = start; i < end; i++) {
            Character ch = s.charAt(i);//返回i的索引的char值
            if (set.contains(ch)) 
            	return false;
            set.add(ch);
        }
        return true;
    }
}
public class LongestSubstring {
	public static void main(String[] args) {
		Solution answer=new Solution();
		System.out.print("The length is "+answer.lengthOfLongestSubstring("ancdafk"));
	}
}

运行结果

The length is 6

代码二:

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();//取得字符串的长度
        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))){//当集合set不包含j索引对应的字符时
                set.add(s.charAt(j++));//就把该字符加入到set中并使j加1
                ans = Math.max(ans, j - i);
            }
            else {
                set.remove(s.charAt(i++));//否则(就是set中包涵了j索引对应的字符),就把最前面的字符(肯定是重复的字符)去掉
            }
        }
        return ans;
    }
}
public class LongestSubstring {
	public static void main(String[] args) {
		Solution answer=new Solution();
		System.out.print("The length is "+answer.lengthOfLongestSubstring("ancdafk"));
	}
}

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转载自blog.csdn.net/weixin_41550144/article/details/89194116
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