哈希表// 无重复字符的最长子串

给定一个字符串,请你找出其中不含有重复字符的 最长子串 的长度。

示例 1:

输入: "abcabcbb"
输出: 3 
解释: 因为无重复字符的最长子串是 "abc",所以其长度为 3。

示例 2:

输入: "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b",所以其长度为 1。

示例 3:

输入: "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke",所以其长度为 3。
     请注意,你的答案必须是 子串 的长度,"pwke" 是一个子序列,不是子串。
class Solution {
    public int lengthOfLongestSubstring(String s) {
        int res = 0, left = 0, right = 0;
        HashSet<Character> t = new HashSet<Character>();
        while(right < s.length()){
            if(!t.contains(s.charAt(right))){
                t.add(s.charAt(right));
                right++;
                res = Math.max(res, t.size());
            }else{
                t.remove(s.charAt(left));
                left++;
            }
        }
        return res;
    }
}
class Solution {
    public int lengthOfLongestSubstring(String s) {
        HashMap<Character,Integer> hashMap = new HashMap<>();
        int res = 0, left = 0;
        for(int i = 0; i < s.length(); i++){
            if(hashMap.containsKey(s.charAt(i))){
                left = Math.max(left, hashMap.get(s.charAt(i)));
            }
            res = Math.max(res,i-left+1);
            hashMap.put(s.charAt(i),i+1);
        }
        return res;
    }
}
class Solution {
    public int lengthOfLongestSubstring(String s) {
        int[] m = new int[256];
        int res = 0, left = 0;
        for(int i = 0; i < s.length(); i++){
            left = Math.max(left, m[s.charAt(i)]);
            res = Math.max(res, i-left+1);
            m[s.charAt(i)] = i+1;
        }
        return res;
    }
}

 

posted @ 2019-01-11 19:59  strawqqhat  阅读(127)  评论(0编辑  收藏  举报
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