leetcode 318. Maximum Product of Word Lengths

曾经终败给现在 2022-08-20 11:18 68阅读 0赞

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