Maximum Product of Word Lengths
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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.
思路:用hashmap来统计,会超时,题目提示,只包含小写字母,这样只有26位,用位移来做,只要两者相与的结果为0,则为互斥关系,更新最大值即可。
public class Solution { public int maxProduct(String[] words) { if(words == null || words.length == 0) return 0; int[] mask = new int[words.length]; for(int i=0; i<words.length; i++){ for(int j=0; j<words[i].length(); j++){ mask[i] |= 1<<(words[i].charAt(j) - 'a'); } } int res = 0; for(int i=0; i<words.length-1; i++){ for(int j=i+1; j<words.length; j++){ if((mask[i] & mask[j])==0) { res = Math.max(res, words[i].length() * words[j].length()); } } } return res; }}
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