leetcode-3. Longest Substring Without Repeating Characters

前言：最近开学事情特别多，隔了一个月没刷题，真是蛋疼。

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

首先想到的最简单的O(n^2)的时间效率代码，提交之后果然超时。

public class Solution {
    public int lengthOfLongestSubstring(String s) {
	int maxLength = 0;
        for(int i = 0 ; i < s.length() - maxLength; i ++){

		HashMap<Character,Integer> s_map = new HashMap<Character,Integer>();
		for(int j = i; j < s.length(); j ++){
			if(s_map.containsKey(s.charAt(j))) break;
			else s_map.put(s.charAt(j),j);
		}
		
		if(s_map.size() > maxLength) maxLength = s_map.size();
	}
	return maxLength;
    }
}

稍微进行了改动，当碰到第一个重复的字符串的时候，将i的值扩大。但是效率还是比较低， You are here! 
Your runtime beats 3.07% of java submissions.。效果有点差，后续再看怎么改进

public class Solution {
    public int lengthOfLongestSubstring(String s) {
	int maxLength = 0;
        for(int i = 0 ; i < s.length() - maxLength; i ++){

		HashMap<Character,Integer> s_map = new HashMap<Character,Integer>();
		for(int j = i; j < s.length(); j ++){
			if(s_map.containsKey(s.charAt(j))) {i = s_map.get(s.charAt(j));break;}
			else s_map.put(s.charAt(j),j);
		}
		
		if(s_map.size() > maxLength) maxLength = s_map.size();
	}
	return maxLength;
    }
}