标签：bsp imu turn rds case 字母 sum char length
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.
含义：扎到两个子串，使得他们长度的乘积最大，要求是一个子串的字符不能在另外一个子串中出现
1 public int maxProduct(String[] words) { 2 // http://blog.csdn.net/li563868273/article/details/51581224 3 // 我们把每个字符串数组看成一个26大小的数组，小写字母a-z是26位，“abcd” 的int值为 0000 0000 0000 0000 0000 0000 0000 1111， 4 // “wxyz” 的int值为 1111 0000 0000 0000 0000 0000 0000 0000，这样两个进行与（&）得到0， 如果有相同的字母则不是0。 5 int len=words.length; 6 if (len<=1) return 0; 7 int[] mask=new int[len]; 8 //abcd可以length=4 9 //words[i].charAt(0) 0左移0位 10 for (int i = 0; i < len; i++) { 11 for (int j = 0; j <words[i].length() ; j++) { 12 mask[i] |= 1<<(words[i].charAt(j)-‘a‘); 13 } 14 } 15 int max= 0; 16 for (int i = 0; i < len; i++) { 17 for (int j = i+1; j <len ; j++) { 18 if((mask[i] & mask[j]) == 0){ 19 max=Math.max(max,words[i].length()*words[j].length()); 20 } 21 } 22 } 23 return max; 24 }
318. Maximum Product of Word Lengths
标签：bsp imu turn rds case 字母 sum char length
原文地址：http://www.cnblogs.com/wzj4858/p/7728126.html
