[leetcode]Longest Substring Without Repeating Characters

Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

算法分析：

思路1：

最土的方法求出所有的子串，继而求出最长的不重复子串。不用试，肯定超时

思路2：

维护一个map，key存储字符，value存储其在s中的下标。当遇到之前出现的字符（下表为j）时，将子串起始下标begin更新为j+1。同时维护maxLength；

代码如下：

public int lengthOfLongestSubstring(String s) {
        int length = s.length();
        Map<Character,Integer> hash = new HashMap<Character,Integer>();
        int currentLength  = 0;
        int maxLength = 0;
        int from = 0;
        for(int i = 0; i < length; i++){
            if(!hash.containsKey(s.charAt(i))){
                currentLength++;
                hash.put(s.charAt(i), i);
            }else{
                if(from <= hash.get(s.charAt(i))){
                    from = hash.get(s.charAt(i)) + 1;
                    currentLength = i - hash.get(s.charAt(i));
                    }else{
                    currentLength++;
                }
                                hash.put(s.charAt(i), i);
            }
            if(currentLength > maxLength){
                maxLength = currentLength;
            }
        }
        return maxLength;
    }

思路2优化：

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0)
            return 0;
        int[] hash = new int[256];
        Arrays.fill(hash, -1);
        int maxLength = 0;
        int pre = -1;
        for (int i = 0; i < s.length(); i++) {
            if (hash[s.charAt(i)] > pre) {
                pre = hash[s.charAt(i)];
            }
            maxLength = Math.max(maxLength, i - pre);
            hash[s.charAt(i)] = i;
        }
        return maxLength;
    }
}