LeetCode 3. Longest Substring Without Repeating Characters

[woodyiiiiiii]

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

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: 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.

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利用滑动窗口法。
定义两个指针代表滑动窗口，两个指针的距离代表定义的字符串长度，并且创建一个HashSet存储字符以便判断字符重复。遍历字符串，如果字符不存在于窗口中，则存入HashSet中，右指针加一；如果存在，左指针加一，并且从HashSet中删除该字符。每次计算更新最长的长度。
注意，如果字符串是“abcdbd”，字符‘b’重复，则左指针加一，把字符'a'剔除了，同时在HashSet中删除了‘a’字符，因为题目要求的是substring，是continuous连续的，所以左指针移动直至删除最先存入的字符‘b’才可。HashSet存的是理想的无重复字符的字符串，窗口长度代表的是HashSet的大小。

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length()== 0)
            return 0;
        HashSet<Character> set = new HashSet<Character>();
        int i =0, j = 0;
        int maxLen = 0;
        while (i < s.length() && j < s.length())
        {
            if(!set.contains(s.charAt(i))) {
                set.add(s.charAt(i));
                i++;
                maxLen = Math.max(maxLen, i-j);
            }
            else {
                set.remove(s.charAt(j));
                j++;
            }
        }
        return maxLen;
    }
}

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参考资料:

• LeetCode原题
• grandyang/leetcode#3

[woodyiiiiiii]

func lengthOfLongestSubstring(s string) int {
        var l = 0
	var r = 0
	n := len(s)
	res := 0
	charMap := make(map[byte]int)
	for r < n {
		c := s[r]
		if _, ok := charMap[c]; ok {
			for l <= r {
				if s[l] == c {
					l++
					break
				}
				delete(charMap, s[l])
				l++
			}
		}
		res = int(math.Max(float64(res), float64(r-l+1)))
		r++
		charMap[c]++
	}
	return res
}