1. 程式人生 > >318. Maximum Product of Word Lengths

318. Maximum Product of Word Lengths

哪些 ive 交集 hat cto abcde add reat urn

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.

解題思路:本題主要是如何快速求出兩個字符串有沒有交集。用一個int值標記每個串a-z哪些字母出現過,之後兩兩比較看是否有交集

class Solution {
public:
    int maxProduct(vector<string>& words) {
        vector<int>v(words.size()+1,0);
        for(int i=0;i<words.size();i++){
            for(int j=0;j<words[i].length();j++){
                v[i]
|=1<<words[i][j]; } } int mmax=0; for(int i=0;i<words.size();i++){ for(int j=i+1;j<words.size();j++){ if(!(v[i]&v[j])&&words[i].length()*words[j].length()>mmax){ mmax=words[i].length()*words[j].length(); } } } return mmax; } };

318. Maximum Product of Word Lengths