leetcode第三题 最长无重复字符子串

 

这题我通过模仿第一题的做法采用hashmap将时间复杂度降至O(n^2),不过leetcode似乎把时间复杂度降到了O(n),或许这就是大佬写的吧。。。。

我的代码

public int lengthOfLongestSubstring(String s) {
		
		if(s.length() == 0) {
			return 0;
		}else if(s.length() == 1){
			return 1;
		}
		
		Map map = new HashMap();
		//int k = 0;
		int[] c = new int[s.length()];
		if(s.length() == 0) {
			return 0;
		}
		for(int i = 0;i < s.length();i++) {
			map.clear();
			for(int j = i;j < s.length();j++) {
				if(map.containsValue(s.charAt(j)) == false) {
				   map.put(j, s.charAt(j));
				   c[i] = j - i + 1 ;
				}else {
				   //c[i] = j - i;	
					break;
				}
			
			}
		
		}
		int k = 0;
		for(int i = 0;i < c.length;i++) {
			if(c[i] > c[k]) {
				k = i;
			}
		}
		return c[k]; 
	}

大佬们的

solution 1

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        Map map = new HashMap<>(); // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 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;
    }
}

solution2 通过使用ASCII实现

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length(), ans = 0;
        int[] index = new int[128]; // current index of character
        // try to extend the range [i, j]
        for (int j = 0, i = 0; j < n; j++) {
            i = Math.max(index[s.charAt(j)], i);
            ans = Math.max(ans, j - i + 1);
            index[s.charAt(j)] = j + 1;
        }
        return ans;
    }
}