LeetCode P318 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.

题目大意：给定一个字符串数组，找到长度乘积最大的两个字符串且这两个字符串中不能存在公共的字符。

思路：

1、既然是找两个长度乘积最大的字符串，那么就可以先对字符串数组进行排序。

2、如何判断两个字符串中不含公共字符？可以使用二进制操作，将字符串转换成一个长度为26的二进制串来代表对应位置的字符是否出现。出现则对应位置1，否则置0，然后将两个字符串相与，结果为0则不含公共字符。

代码如下：

public class Solution {    public int maxProduct(String[] words) {        int max = 0;        if (words.length <= 1) {            return 0;        }        //排序，逆序        Arrays.sort(words, new Comparator<String>() {            @Override            public int compare(String o1, String o2) {                return o2.length() - o1.length();            }        });        for (int i = 0; i < words.length-1; i++) {            //因为是逆序，所以一旦出现如下情况，后面的乘积都小于max            if (words[i].length() * words[i].length() <= max)break;            for (int j = i+1; j < words.length; j++) {                if (!hasSameChar(words[i], words[j])) {                    int tmp = words[i].length() * words[j].length();                    max = Math.max(max, tmp);                    break;//后面的情况只会更小，故跳出                }            }        }        return max;    }    //判断两个字符串中是否含有公共字符    private boolean hasSameChar(String s1, String s2){        int one = 0;        int two = 0;        for (int i = 0; i < s1.length(); i++) {            one |= (1 << (s1.charAt(i) - 'a'));        }        for (int i = 0; i < s2.length(); i++) {            two |= (1 << (s2.charAt(i) - 'a'));        }        one = one & two;        return !(one == 0);    }}