leetcode [318]Maximum Product of Word Lengths

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0.

Example 1:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16 
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4 
Explanation: The two words can be "ab", "cd".

Example 3:

Input: ["a","aa","aaa","aaaa"]
Output: 0 
Explanation: No such pair of words.

题目大意：

给定一个字符串组，找到两个字符串满足两个字符串没有公共的部分，并且两个字符串长度的乘积是最大的。

解法：

使用了一个flag数组来记录字符串中出现的字符，如果说两个flag[i]&flag[j]==0，说明两个字符串没有公共的部分，满足题目要求。

java:

class Solution {
    public int maxProduct(String[] words) {
        int n=words.length;
        int[] flag=new int[n];
        for (int i=0;i<n;i++){
            for (int j=0;j<words[i].length();j++){
                flag[i]|=1<<(words[i].charAt(j)-'0');
            }
        }
        int res=0;
        for (int i=0;i<n;i++){
            for (int j=i+1;j<n;j++){
                if ((flag[i]&flag[j])==0) res=Math.max(res,words[i].length()*words[j].length());
            }
        }

        return res;
    }
}