[leetcode]318. Maximum Product of Word Lengths单词长度最大乘积

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:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4
Explanation: The two words can be "ab", "cd".

题意：

给定一堆单词，要求找出俩单词长度的最大乘积，要求俩单词不能有相同字母。

思路：

判断noOverLapping: 用2个set分别标记俩单词， 扫一遍set，若发现某个字母被同时标记过，则有重叠。

取最大乘积长度：两重for循环，两两比较，更新最大值。

代码：

 class Solution {
     public int maxProduct(String[] words) {
         int result = 0;
         for (int i = 0; i < words.length; ++i) {
             for (int j = i + 1; j < words.length; ++j) {
                 int tmp = words[i].length() * words[j].length();
                 if ( noOverLapping(words[i], words[j])&& tmp > result) {
                     result = tmp;
                 }
             }
         }
         return result;
 }
     private boolean noOverLapping(String a , String b){
         boolean[] setA = new boolean[256];
         boolean[] setB = new boolean[256];

         for(int i = 0; i < a.length(); i++){
             setA[a.charAt(i)] = true;
         }

         for(int i = 0; i < b.length(); i++){
             setB[b.charAt(i)] = true;
         }

         for(int i = 0; i < 256; i++){
             if(setA[i] == true && setB[i] == true){
                 return false;
             }
         }

         return true;
     }
 }

可以进一步优化：对于如何判断俩单词有没有相同字母，可用位向量表示每个字母是否出现即可，俩位向量异或即可得出是否有相同字母。

 public class Solution {
     public int maxProduct(String[] words) {
         final int n = words.length;
         final int[] hashset = new int[n];

         for (int i = 0; i < n; ++i) {
             for (int j = 0; j < words[i].length(); ++j) {
                 hashset[i] |= 1 << (words[i].charAt(j) - 'a');
             }
         }

         int result = 0;
         for (int i = 0; i < n; ++i) {
             for (int j = i + 1; j < n; ++j) {
                 int tmp = words[i].length() * words[j].length();
                 if ((hashset[i] & hashset[j]) == 0 && tmp > result) {
                     result = tmp;
                 }
             }
         }
         return result;
     }
 }