318. Maximum Product of Word Lengths

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本文介绍了一种高效算法，用于找出两个不包含相同字母的字符串，以求得它们长度乘积的最大值。通过使用位运算预处理字符串，极大提高了查找速度。

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:

Given ["abcw", "baz", "foo", "bar", "xtfn", "abcdef"]
Return 16
The two words can be "abcw", "xtfn".

Example 2:

Given ["a", "ab", "abc", "d", "cd", "bcd", "abcd"]
Return 4
The two words can be "ab", "cd".

Example 3:

Given ["a", "aa", "aaa", "aaaa"]
Return 0

No such pair of words.

基本思路就是两两比较，但是之前肯定要对string做一些预处理

/*
 * 用个数组表示每个string，通过比较每个数组有没有重叠来判断有没有相同的字符
 * 转念一想，用什么数组，一个integer足矣
 */
class Solution {
    public int maxProduct(String[] words) {
        int[] a = new int[words.length];
        for(int i=0; i<words.length; i++) {
        	for(char c : words[i].toCharArray())
        		a[i] |= (1<<(c-'a'));
        }
        
        int max = 0;
        for(int i=0; i<a.length; i++)
        	for(int j=i+1; j<a.length; j++)
        		if((a[i] & a[j]) == 0)
        			max = Math.max(max, words[i].length() * words[j].length());
        
        return max;
    }
}