leetcode 318. 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 0
No such pair of words.

bool Compare(const string &s1, const string& s2){return s1.length() > s2.length();}class Solution {public:int maxProduct(vector<string>& words) {if (words.size() < 2)return 0;sort(words.begin(), words.end(), Compare);int max = 0;for (int i = 0; i < words.size() - 1; i++)for (int j = i + 1; j < words.size(); j++){//set<char>intersec;  //set_intersection(aa[i].begin(), aa[i].end(), aa[j].begin(), aa[j].end(), inserter(intersec, intersec.begin()));  int t = words[i].size()*words[j].size();if (t > max){bool flag = true;int a[26] = { 0 }, b[26] = { 0 };for (int h = 0; h < words[i].length(); h++)a[words[i][h] - 'a'] = 1;for (int h = 0; h < words[j].length(); h++)b[words[j][h] - 'a'] = 1;for (int h = 0; h < 26; h++)if (a[h] == 1 && b[h] == 1){flag = false;break;}if (flag)max = t;}if (i + 2 < words.size() && words[i + 1].length()*words[i + 2].length() <= max)return max;}return max;}};

accepted