LeetCode|99% ——Longest Substring Without Repeating Characters

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题目：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.

也就是求一个给定字符串的最长连续不重复子串。首先还是上最直观也是最笨的两层循环解法：

class Solution {
    public int lengthOfLongestSubstring(String s) {

        int maxLength = 0;
        for (int i = 0; i < s.length(); i++) {
            Set<Character> set = new HashSet<Character>();
            for (int j = i; j < s.length(); j++) {
                if (set.contains(s.charAt(j))) {

                    break;
                } else {
                    set.add(s.charAt(j));
                    maxLength = set.size() > maxLength ? set.size() : maxLength;
                }
            }
        }

        return maxLength;

    }
}

该解法耗时159ms，排在5.22%，显然是很差的。

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        Map<Character, Integer> map = new HashMap<Character, Integer>(128);
        int max = 0, j = 0;
        for (int i = 0; i < s.length(); i++) {
            if (map.containsKey(s.charAt(i)))
                j = Math.max(j, map.get(s.charAt(i)));
            map.put(s.charAt(i), i + 1);
            max = Math.max(max, i - j + 1);
        }
        return max;
    }
}

O(n)解法。耗时62ms排在54.01%。

    public class Solution {
    public int lengthOfLongestSubstring(String s) {
            int[] map = new int[128];
    int max = 0, j = 0;
    char[] str = s.toCharArray();
    int length = s.length();

    for(int i = 0; i < length; i++) {
        if(map[str[i]] > 0)
            j =  Math.max(j, map[str[i]]);
        map[str[i]] = i + 1;
        max = Math.max(max, i - j + 1);
    }
    return max;
    }
}

用数组替代HashMap和String。耗时41ms。排在99.25%。