Longest-Substring-Without-Repeating-Characters，

Longest-Substring-Without-Repeating-Characters，

---

title: Longest Substring Without Repeating Characters
date: 2017-10-24 22:58:49
categories:
- algorithm
tags:

- algorithm

摘要：Brute Force

正文:

Approach #1 Brute Force [Time Limit Exceeded]

第一眼看到 求最长的非重复子串，就是三个for循环 求解，去掉非法的判断，花了十分钟得到了一个不得使用“蛮力”，时间复杂度O（n*3）可是明明用了hashSet的Contain（）方法了，Exception代码如下:

class Solution {
        public static boolean allUnique(String string,int start,int end){

        Set<Character> set =new HashSet<Character>();
        for (int i = start; i < end; i++){
            if (set.contains(string.toCharArray()[i])){
                return false;
            }
            set.add(string.toCharArray()[i]);
        }

        return true;
    }
    public int lengthOfLongestSubstring(String s) {
        int length = s.length();
        int ans = 0;
        for (int i = 0; i < length; i++){
            for (int j = i + 1; j <= length; j++){
                if (allUnique(s, i, j)){
                    ans = Math.max(ans, j - i);
                }
            }
        }
        return ans;
    }

}

看了别人通过的代码才知道我的暴力破除的确不够优雅啊 ，奉上正确的代码，很简单就不注释了

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