318. Maximum Product of Word Lengths | 字符串长度相乘最大值

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本文介绍了一种算法，用于找出字符串数组中两个没有共同字母的字符串长度乘积的最大值。通过使用位操作来快速判断两个字符串是否有共同字母，并以此为基础遍历整个数组寻找最优解。

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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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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思路：创建一个int数组，长度为string数组的长度，其中每个数对应该字符串中对应位置的字符串，值是26位，最低位为a对应，最高位对应z，当该字母存在时，则对应位位1，体现在代码中则为 value[i] |= 1 << (tmp.charAt(j) - 'a');随后再遍历字符串数组，当两个字符串对应的value值相与为0,(若存在相同的字符，则相与会大于0)且长度相乘大于当前最大值，遍历完返回最大值即可。

public class Solution {
    	public int maxProduct(String[] words) {

		if (words.length == 0 || words.length == 1) {
			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 - 1; 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;
	}
}