LeetCode 318. Maximum Product of Word Lengths（最大單詞長度乘積）

原題網址：https://leetcode.com/problems/maximum-product-of-word-lengths/

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 

No such pair of words.

方法：可以通過直方圖統計來比較兩個單詞是否有共同的字母，但有個更快捷的方法就是使用26個比特來比較。

public class Solution {
    private int hash(char[] wa) {
        int hash = 0;
        for(int i=0; i<wa.length; i++) {
            hash |= (1<<(wa[i]-'a'));
        }
        return hash;
    }    
    public int maxProduct(String[] words) {
        char[][] was = new char[words.length][];
        int[] hashes = new int[words.length];
        for(int i=0; i<words.length; i++) {
            was[i] = words[i].toCharArray();
            hashes[i] = hash(was[i]);
        }
        int max = 0;
        for(int i=0; i<words.length-1; i++) {
            for(int j=i+1; j<words.length; j++) {
                if ((hashes[i] & hashes[j]) == 0) {
                    int m = was[i].length * was[j].length;
                    if (m > max) max = m;
                }
            }
        }
        return max;
    }
}
           

方法二：使用比特進行哈希。

public class Solution {
    public int maxProduct(String[] words) {
        int[] f = new int[words.length];
        for(int i=0; i<words.length; i++) {
            char[] wa = words[i].toCharArray();
            for(int j=0; j<wa.length; j++) {
                f[i] |= 1 << (wa[j]-'a');
            }
        }
        int max = 0;
        for(int i=0; i<words.length-1; i++) {
            for(int j=i+1; j<words.length; j++) {
                if ((f[i] & f[j]) == 0) max = Math.max(max, words[i].length() * words[j].length());
            }
        }
        return max;
    }
}