【Leetcode】：318. Maximum Product of Word Lengths 问题 in JAVA

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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.

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这道题没思路，参考这位仁兄：http://www.jianshu.com/p/bb84b0f866c9

主要难点在于如何简单的判断两个word是否存在相同字母，方法如下：小写字母一共有26个，一个int是32位，那么可以用int的每个位来代表一个字母

例如：abcd = 0000 0000 0000 0000 0000 0000 0000 1111；

例如：abcg = 0000 0000 0000 0000 0000 0000 0100 0111；

那么对两个word生成的两个int做相与，如果结果为0那么一定表示没有相同的字母

public class Solution {    public int maxProduct(String[] words) {        if (words.length <= 1) {            return 0;        }        int[] preProcessed = new int[words.length];        for (int i = 0; i < words.length; i++) {            for (int j = 0; j < words[i].length(); j++) {                preProcessed[i] |= 1 << (words[i].charAt(j) - 'a');            }        }        int maxL = 0;        for (int i = 0; i < words.length; i++) {            for (int j = i+1; j < words.length; j++) {                if ((preProcessed[i] & preProcessed[j]) == 0) {//没有重复字母                    int product = words[i].length() * words[j].length();                    if (maxL < product) {                        maxL = product;                    }                }            }        }        return maxL;    }}

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