【LeetCode】318. 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.

class Solution {public:    int maxProduct(vector<string>& words) {        int length=words.size();        int * wordsMap= new int[length];                memset(wordsMap, 0, sizeof(int)*length);                for(int i=0;i<length;i++){            for(int j=0;j<words[i].size();j++){                wordsMap[i] = wordsMap[i] | (0x00000001<<(words[i].at(j)-'a'));            }        }                int maxSum=0;                for(int i=0;i<length;i++){            for(int j=i+1;j<length;j++){                if((wordsMap[i] & wordsMap[j])==0){                    if(words[i].size()*words[j].size()>maxSum){                        maxSum=words[i].size()*words[j].size();                    }                }            }        }        return maxSum;    }};

心得体会：
万万没想到这个代码还是挺快的

思想就是将每一个字符串换成对应的32位Int.

因为只有26个字母。所以出现的字母在32位的int中的一位标志着这个字字母有没有出现过。

最后再用一个O(n^2)的循环，比对这些int是否有相同的位，如果有就是出现了同样的字母

该逻辑用逻辑与实现

预处理是O(n*m) m为字符串长度

判断是O(n^2)

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