Leetcode: Maximum Product of Word Lengths

最新推荐文章于 2025-04-19 00:25:33 发布

转载最新推荐文章于 2025-04-19 00:25:33 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/EdwardLiu/p/5094103.html

本文介绍了一种使用位操作优化字符串匹配算法的方法，该方法应用于寻找两个不含相同字母的最大长度乘积的问题。通过使用位掩码减少重复计算并采用适当的剪枝策略，实现了高效的算法解决方案。

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.

想到String的char 查重，最好的方法就是用bit manipulation

注意O(N^2)的iteration里面可以适当pruning

 1 public class Solution {
 2     public int maxProduct(String[] words) {
 3         if (words==null || words.length==0) return 0;
 4         int max = 0;
 5         int[] bitmasks = new int[words.length]; 
 6         Arrays.sort(words, new Comparator<String>(){
 7             public int compare(String s1, String s2) {
 8                 return s2.length()-s1.length();
 9             }
10         });
11         
12         for (int i=0; i<words.length; i++) {
13             for (int j=0; j<words[i].length(); j++) {
14                 bitmasks[i] |= 1<<((int)(words[i].charAt(j) - 'a'));
15             }
16         }
17         
18         for (int i=0; i<words.length-1; i++) {
19             for (int j=i+1; j<words.length; j++) {
20                 if (words[i].length() * words[j].length() <= max) break; //pruning
21                 if ((bitmasks[i] & bitmasks[j]) != 0) continue;
22                 max = words[i].length() * words[j].length();
23                 break;//pruning
24             }
25         }
26         return max;
27     }
28 }

转载于:https://www.cnblogs.com/EdwardLiu/p/5094103.html