LeetCode 318. Maximum Product of Word Lengths

Posted 2020-11-20 wolfpack

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:

Input: ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16 
Explanation: The two words can be "abcw", "xtfn".

Example 2:

Input: ["a","ab","abc","d","cd","bcd","abcd"]
Output: 4 
Explanation: The two words can be "ab", "cd".

Example 3:

Input: ["a","aa","aaa","aaaa"]
Output: 0 
Explanation: No such pair of words.

 

思路:

1.需要找到有效的方法来比较字符串中是否含有相同的元素。 (bitmask)

2.找到这种方法之后，直接遍历整个数组找最大的乘积就得出解了。(brute force)

Code：

 1 #include <vector>
 2 #include <string>
 3 using namespace std;
 4 int maxProduct(vector<string>& words) {
 5     int res = 0, n = words.size();
 6     vector<int> charBits(n, 0);
 7     for (int i = 0; i < n; i++) {
 8         charBits[i] = 0;
 9         for (int j = 0; j < words[i].size(); j++) {
10             charBits[i] |= 1 << (words[i][j] - ‘a‘);
11         }
12     }
13 
14     for (int i = 0; i < n; i++) {
15         for (int j = 0; j < n; j++) {
16             if ((charBits[i] & charBits[j]) == 0) {
17                 int product = words[i].size() * words[j].size();
18                 if (res < product) {
19                     res = product;
20                 }
21             }
22         }
23     }
24     return res;
25 }

详细解释：

Part1:

第10行代码：

//bitmask: bit operation,move "0x0001" left "(words[i][j] - ‘a‘)" bits; this way we can store different characters in charBits[i]

charBits[i] |= 1 << (words[i][j] - ‘a‘);

从‘a‘到‘z‘一共有26个字母，一个int变量有32bit，每个字母与‘a‘相减得到的值（即0x0001需要左移的位置）不会超出int变量的范围。

通过将0x0001（即1）左移 “ words[i][j] - ‘a‘ ”位，我们将字母words[i][j]的信息存储在了位操作后的int数字上。（注意：即使存在重复的字母，也不会有影响。因为我们要存储的不是字符串的全部信息，仅仅是单词中所有不同字母的信息）

然后再通过 “|=”操作，将单词words[i]的j位置的字母words[i][j]存到变量charBits[i]（即bitmask）中。

遍历整个单词，即可得到整个字符串的bitmask：charBits[i]。

遍历完整个words之后，我们就得到了每个单词words[i]的bitmask。

Part2:

之后在第16行代码中：

           if ((charBits[i] & charBits[j]) == 0) {

将不同单词words[i]和words[j]的bitmask（charBits[i]和charBits[j]）,作&运算，只要不等于0，就说明两个单词有相同的字母。

一旦(charBits[i] & charBits[j]) == 0，则说明两个单词没有任何相同的字母。

 

---

 

延伸：

1.题设中的条件是只有小写字母，如果有大写字母怎么办？

答：用long或者bit[64]来存储字母的信息。

char c = words[i][j]
if (islower(c) {
    charBits[i] |= (long)1 << ( c - ‘a‘);
} else {
    charBits[i] |= (long)1 << (( c - ‘A‘)+32);
}