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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.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
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1 public class Solution { 2 public int maxProduct(String[] words) { 3 int ans = 0; 4 int[] bitmap = new int[words.length]; 5 6 for(int i = 0; i < words.length; i++){ 7 for(int j = 0; j < words[i].length(); j++){ 8 bitmap[i] |= 1 << (words[i].charAt(j) - ‘a‘); 9 } 10 } 11 12 for(int i = 0; i < bitmap.length; i++){ 13 for(int j = i+1; j < bitmap.length; j++){ 14 if((bitmap[i] & bitmap[j]) == 0 && words[i].length()*words[j].length() > ans){ 15 ans = words[i].length()*words[j].length(); 16 } 17 } 18 } 19 20 return ans; 21 } 22 }
之前都没有做过类似的题目，题目要找出字符串数组中两个 没有一个字母相同的字符串乘积的最大值，学习了大神的做法。
字母只有26个，java 中一个int 有32位。
通过 bitmap[i] |= 1 << (words[i].charAt(j) - ‘a‘); 即可记录下下标为i的字符串字符特征。
全部扫描一遍，即可记录所有的字符串特征， 之后二重循环即可得答案。
318. Maximum Product of Word Lengths java solutions
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原文地址：http://www.cnblogs.com/guoguolan/p/5608102.html
