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:

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.

哎，有想把字符串转换成别的形式，方便处理。。。但还是没想到，好菜啊

用了位处理

public class Solution {
    public static int maxProduct(String[] words) {
	if (words == null || words.length == 0)
		return 0;
	int len = words.length;
	int[] value = new int[len];
	for (int i = 0; i < len; i++) {
		String tmp = words[i];
		value[i] = 0;
		for (int j = 0; j < tmp.length(); j++) {
			value[i] |= 1 << (tmp.charAt(j) - 'a');
		}
	}
	int maxProduct = 0;
	for (int i = 0; i < len; i++)
		for (int j = i + 1; j < len; j++) {
			if ((value[i] & value[j]) == 0 && (words[i].length() * words[j].length() > maxProduct))
				maxProduct = words[i].length() * words[j].length();
		}
	return maxProduct;
}
}