[LeetCode]3.无重复字符的最长子串

自己的解法

将字符串转化为字符数组，从首字符向后扫描字串，找到不重复的最长字串

class Solution {
    public int lengthOfLongestSubstring(String s) {
        char[] chars = s.toCharArray();
        HashSet<Character> hs = new HashSet();
        int size = 0;
        for(int j = 0; j < chars.length; j++){
            for(int i = j; i < chars.length; i++){
                if(hs.contains(chars[i])){
                        break;
                }
                hs.add(chars[i]);
            }
            if(size < hs.size()){
                size = hs.size();
            }
            hs.clear();
        }
        return size;
    }
}

滑动窗口

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

优化滑动窗口

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

假设字符集为 ASCII 128

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int[] index = new int[128];
        int n = s.length();
        int ans = 0;
        for(int i=0, j = 0; j < n; j++){
            i = Math.max(index[s.charAt(j)],i);
            ans = Math.max(ans,j-i+1);
            index[s.charAt(j)] = j+1;
        }
        return ans;
    }
}