Problem 318: Maximum Product of Word Lengths
https://leetcode.com/problems/maximum-product-of-word-lengths/?tab=Description
思路
这道题很有意思。解法里用
count[i]
来表明对应的word[i]
的每一位上的字母的情况。这里的word[i]
是用二进制表示的,从右向左代表abcde...xyz
这几个数最后循环一遍以后,判断最大值是多少。这里 check 的时候用
&
来判断两个数是否是“没有 share 的元素”
public class Solution {
public int maxProduct(String[] words) {
if (words == null || words.length == 0) return 0;
int len = words.length;
int[] count = new int[len];
for (int i = 0; i < len; i++) {
String word = words[i];
count[i] = 0;
for (int j = 0; j < word.length(); j++) {
count[i] |= 1 << (word.charAt(j) - 'a');
}
}
int product = 0;
for (int i = 0; i < len; i++) {
for (int j = i + 1; j < len; j++) {
if ((count[i] & count[j]) == 0 && words[i].length() * words[j].length() > product) {
product = words[i].length() * words[j].length();
}
}
}
return product;
}
}
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