LeetCode:Maximum Product of Word Lengths

Maximum Product of Word Lengths

Total Accepted: 5729 Total Submissions: 15214 Difficulty: Medium

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.

Hide Tags

 Bit Manipulation

思路：

1.预处理(1)先将字符串数组words以长度从大到小排序；

2.预处理(2)将words中的每个字符串(word)中含有的字母用分别用整数来保存（保存方法见代码）；

3.求product(结果)：两层loop，两两比较，用&运算对2中的整数进行比较（判断是否有相同字母）；

其中还有两个可“剪枝”技巧。

code:

class Solution {public:    int maxProduct(vector<string>& words) {        int len = words.size();        if(0==len) return 0;        sort(words.begin(), words.end(), cmp); // 【1】        vector<int> bits(len,0);        for(int i=0;i<len; i++)            for(int j=0;j<words[i].size();j++){                bits[i] |= 1 << (words[i][j]-'a'); // 【2】            }        int maxSqu = 0;        for(int i=0;i<len; i++) {  // 【3】            if(words[i].size() * words[i].size() <= maxSqu) break; // 剪枝            for(int j=i+1;j<len;j++) {                if(!(bits[i] & bits[j])) {                    int tmp = words[i].size() * words[j].size();                    maxSqu = tmp > maxSqu ? tmp:maxSqu;                    break; // 剪枝                }            }        }        return maxSqu;    }    static bool cmp(string a, string b) {        return a.size() > b.size();    }};