LeetCode Maximum Product of Word Lengths

原题链接在这里：https://leetcode.com/problems/maximum-product-of-word-lengths/

用mark[i]数组来标记 words[i]都含有哪些字母，假若有个'c', 就在第三位上有个1. 

通过mark[i] & mark[j] 是否为0来确定是否有相同的字母。

排序words是为了能尽量节省一些时间，去掉没有意义的比较。

Time Complexity: O(n^2), n是有多少个string.

Space: O(n), 生成了mark.

AC Java:

public class Solution {
    public int maxProduct(String[] words) {
        if(words == null || words.length == 0){
            return 0;
        }
        
        //排序words, 按照长度大到小, 为了后面节省时间
        Arrays.sort(words, new Comparator<String>(){
            public int compare(String s1, String s2){
                return s2.length() - s1.length();
            }
        });
        
        int [] mark = new int[words.length];
        for(int i = 0; i<words.length; i++){
            for(int j = 0; j<words[i].length(); j++){
                mark[i] |= (1 << words[i].charAt(j)-'a');
            }
        }
        
        int res = 0;
        for(int i = 0; i<words.length; i++){
            if(words[i].length() * words[i].length() <= res){
                //因为words已经按照长度大到小排序了，若是words[i]长度平方还小于res, 后面的更小
                break;
            }
            for(int j = i+1; j<words.length; j++){
                if((mark[i] & mark[j]) == 0){
                    res = Math.max(res, words[i].length() * words[j].length());
                    //后面更小，没有必要继续了
                    break;
                }
            }
        }
        
        return res;
    }
}