[LEETCODE]318. Maximum Product of Word Lengths

[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.

这套题用的是位运算。因为int 比a~z 的26位要多，所以我们可以用不同数位来表示不同的字母。然后做与运算（注意这里是& 而不是 &&），然后如果结果是0 说明两个字符串中没有相同的字母。最后我们算出这些字符串中最大长度乘积。

class Solution {
public:
int maxProduct(vector<string>& words) {
int size = words.size();
if (size < 2) return 0;
vector<int>len(size);
vector<int>nums(size);
for (int i = 0; i < size; ++i) {
len[i] = words[i].size();
nums[i] = 0;
for (int j = 0; j < len[i]; ++j) {
nums[i] |= 1 << (words[i][j] - 'a');
}
}
int max = 0;
for (int i = 0; i < size; i++) {
for (int j = i + 1; j < size; j++) {
if ((nums[i] & nums[j]) == 0) {
int temp = len[i] * len[j];
if (temp > max) max = temp;
}
}
}
return max;
}
};