“Leetcode是一个强大的OJ网站，很多公司的面试题目都可以在这里找到”
题目
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.
翻译
给定一个string数组words，找出没有重复字母的最大length(word[i]) * length(word[j])。
你可以假设每个单词只有小写字母，如果没有这样的两个单词，返回0.
Tags
Bit Manipulation
代码
Java:
public int maxProduct(String[] words) {
int max = 0;
int length = words.length;
int[] wordList = new int[length];
for(int i = 0; i < length; i++)
for(char chr: words[i].toCharArray())
wordList[i] |= 1 << (int)(chr - 'a');
for(int i = 0; i < length; i++){
for(int j = i + 1; j < length; j++){
if((wordList[i] & wordList[j]) == 0)
max = Math.max(max, words[i].length() * words[j].length());
}
}
return max;
}
讲解
int类型有32位，小写字母只有26位，所以可以用一个int型存储所有小写字母，然后通过与操作可以知道两个
单词是否有重复，进而可以算出长度积的最大值。
习题地址
Maximum Product of Word Lengths
