leetcode Maximum Product of Word Lengths

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本文介绍了一种使用位运算解决LeetCode上一道题目——最大单词长度乘积的方法。该题要求找出字符串数组中不包含相同字母的两单词的最大长度乘积。文章通过示例解释了如何利用整数存储每个单词的字母信息，并通过位运算快速判断两单词是否包含相同的字母。

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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.

Follow up:

Could you do better than O(n2), where n is the number of words?

转自我的新博客：细语呢喃 > leetcode Maximum Product of Word Lengths

其实因为全部都是小写的字母，用int 就可以存储每一位的信息。这就是位运算

elements[i] |= 1 << (words[i][j] – ‘a’); //把words[i][j] 在26字母中的出现的次序变为1
elements[i] & elements[j] // 判断是否有交集只需要两个数按位与（AND）运算即可

class Solution {
public:
	int maxProduct(vector<string>& words) {
		int n = words.size();
		vector<int> elements(n, 0);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < words[i].length(); j++)
				elements[i] |= 1 << (words[i][j] - 'a');
		}
		int ans = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (!(elements[i] & elements[j]) && words[i].length() * words[j].length()  > ans)
					ans = words[i].length() * words[j].length();
			}
		}
		return ans;
	}
};

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