LeetCode-318. 最大单词长度乘积

给定一个字符串数组 words，找到 length(word[i]) * length(word[j]) 的最大值，并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词，返回 0。

示例 1:

输入: ["abcw","baz","foo","bar","xtfn","abcdef"]
输出: 16 
解释: 这两个单词为 "abcw", "xtfn"。
示例 2:

输入: ["a","ab","abc","d","cd","bcd","abcd"]
输出: 4 
解释: 这两个单词为 "ab", "cd"。
示例 3:

输入: ["a","aa","aaa","aaaa"]
输出: 0 
解释: 不存在这样的两个单词。

附上作者原文地址：

https://leetcode-cn.com/problems/maximum-product-of-word-lengths/solution/c-jian-dan-wei-yun-suan-bu-guo-huan-ke-yi-yong-bin/

思路：

利用位运算，将字符串转化成为二进制进行判断是否拥有重复的子字符。

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#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
	/* 将字符串转化为二进制 */
	int str2int(const string& s) {
		int res = 0;
		for (auto c : s)
			res |= 1 << (c - 'a');
		return res;
	}

	int maxProduct(vector<string>& words) {
		int N = words.size();
		vector<int> m(N, 0);
		for (int i = 0; i < N; ++i)
			m[i] = str2int(words[i]);

		/* 展示生产的二进制 */
		for (int i = 0; i < N; ++i) {
			cout << "m[" << i << "]:";
			ToBinary(m[i]);
		}

		int res = 0;
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < i; ++j) {
				if ((m[i] & m[j]) == 0 && words[i].size() * words[j].size() > res)
					res = words[i].size() * words[j].size();
			}
		}
		return res;
	}

private:
	void ToBinary(int num) {
		int BinaryArr[100];
		memset(BinaryArr, 0x00, sizeof(BinaryArr));
		int t = 0;
		while (num) {
			BinaryArr[t++] = num & 1;
			num = num >> 1;
		}
		for (int i = t - 1; i >= 0; i--) {
			printf("%d", BinaryArr[i]);
		}
		printf("\n");
	}

};

int main() {
	vector<string> test1 = { "a","aa","aaa","aaaa" };
	vector<string> test2 = { "abcw", "baz", "foo", "bar", "xtfn", "abcdef" };
	Solution* ps = new Solution();

	cout << "res:" << ps->maxProduct(test2) << endl;

	return 0;
}