(LeetCode) 3、Longest Substring Without Repeating Characters

3. Longest Substring Without Repeating Characters

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.

思路：

从位置begin开始，一直找到end, str[end+1]为[begin, end]中的某个字符重复，即从begin开始，最大不重复字符子串长度为end - begin;

判断字符是否存在某个字符集中，暂时只想到 set

代码：

public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        int ans = 0;
        Set<Character> set = new HashSet<Character>();
		for(int i = 0; i < n; i++){
			if(n - i < ans) break; //剩余字符没有ans大，不需计算
			set.clear();
			for(int j = i; j < n; j++){
				char c = s.charAt(j);
				if(set.contains(c)){
					ans = Math.max(ans, j - i);
					break;
				}
				else
					set.add(c);
			}
			ans = Math.max(ans, set.size());
		}
        return ans;
    }

 优化：

假设[x, y]为从x开始的最大无重复字符子串，str[y+1]与[x, y]中的某个字符重复，可以肯定从[x+1, y]是没有重复字符的，如果有的话，在[x, y]的过程中就会出现，

所以从x+1位置开始，只需要set.remove(str[x]), 从y+1开始计算，set中的值就没有必要clear()

代码：

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int n = s.length();
        int ans = 0;
        int pre = 0;
        Set<Character> set = new HashSet<Character>();
		for(int i = 0; i < n; i++){
			if(n - i < ans) break; //剩余字符没有ans大，不需计算
			if(i > 0) set.remove(s.charAt(i-1));
			while(pre < n){
				char c = s.charAt(pre);
				if(set.contains(c))
					break;				
				else
					set.add(c);
				pre++;
			}
			ans = Math.max(ans, set.size());
		}
        return ans;
    }
}