java面试题（10）

原题：

// 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.

解析：判断两个字符没有相同的字母，且个数相乘最大

答案：

public class Solution {

public int maxProduct(String[] words) {

if(words.length == 0 || words == null) return 0;

int length = words.length;
int[] value = new int[length];
int max = 0;

for(int i = 0; i < length; i++) {

String temp = words[i];

value[i] = 0;

for(int j = 0; j < temp.length(); j++) {

value[i] |= 1 << (temp.charAt(j) - 'a');

978e
}

}

for(int i = 0; i < length; i++) {

for(int j = 1; j < length; j++) {

if((value[i] & value[j]) == 0 && (words[i].length() * words[j].length()) > max) {

max = words[i].length() * words[j].length();

}

}

}

return max;

}

}

解析：主要利用二进制的特点

一个字符全部按位或之后，如果两个字符按位与，如果有相同的字母，则按位与肯定不为0，因为他们肯定有相同的1，按位与后会有1出现