leetcode 318. Maximum Product of Word Lengths

题目描述

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.

解题思路

本题感觉难度应该是easy，不知道为什么排在medium，首先AC很简单，做一个嵌套for循环就OK，这道题的关键在于用空间换取时间，注意到题目中对单词有一个要求，就是说两个单词不能有相同的单词，如果我们用普通方法去做，每做一次判断就要循环遍历每两个string作比较，很显然多算了很多次，我们的想法自然是能否加速比较过程，这里的思路就比较新颖了，判断是否两个元素是否相同通常会想到相与，而对于字母元素来说，最多26个，int型的变量足够保存一个string元素包含的所有不同元素，因此，我们每次只用做一个与操作就可以判断两个string是否有元素重复。具体实现如下：

int maxProduct(vector<string>& words) {
        if(words.empty())
        return 0;
        vector<int> bit(words.size(),0);
        int i = 0;
        for(auto &val:words) {
            for(auto&var:val) {
                bit[i]|=(1<<(var-'a'));
            }
            i++;
        }
        int maxVal = 0;
        for(int m = 0;m<words.size();++m)
            for(int n = m+1;n<words.size();++n) {
                if((bit[m]&bit[n])==0&&maxVal<words[m].size()*words[n].size()) {
                    maxVal = words[m].size()*words[n].size();
                }
            }
        return maxVal;

    }