318. 最大单词长度乘积

English Version

题目描述

给你一个字符串数组 words ，找出并返回 length(words[i]) * length(words[j]) 的最大值，并且这两个单词不含有公共字母。如果不存在这样的两个单词，返回 0 。

 

示例 1：

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

示例 2：

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

 

提示：

• 2 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] 仅包含小写字母

解法

Python3

class Solution:
    def maxProduct(self, words: List[str]) -> int:
        n = len(words)
        mask = [0] * n
        for i, word in enumerate(words):
            for ch in word:
                mask[i] |= 1 << (ord(ch) - ord('a'))
        ans = 0
        for i in range(n - 1):
            for j in range(i + 1, n):
                if mask[i] & mask[j] == 0:
                    ans = max(ans, len(words[i]) * len(words[j]))
        return ans

Java

class Solution {
    public int maxProduct(String[] words) {
        int n = words.length;
        int[] masks = new int[n];
        for (int i = 0; i < n; ++i) {
            for (char c : words[i].toCharArray()) {
                masks[i] |= (1 << (c - 'a'));
            }
        }
        int ans = 0;
        for (int i = 0; i < n - 1; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if ((masks[i] & masks[j]) == 0) {
                    ans = Math.max(ans, words[i].length() * words[j].length());
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int n = words.size();
        vector<int> mask(n);
        for (int i = 0; i < n; ++i)
            for (char ch : words[i])
                mask[i] |= 1 << (ch - 'a');
        int ans = 0;
        for (int i = 0; i < n - 1; ++i)
            for (int j = i + 1; j < n; ++j)
                if (!(mask[i] & mask[j]))
                    ans = max(ans, (int)(words[i].size() * words[j].size()));
        return ans;
    }
};

Go

func maxProduct(words []string) int {
	n := len(words)
	mask := make([]int, n)
	for i, word := range words {
		for _, c := range word {
			mask[i] |= (1 << (c - 'a'))
		}
	}
	ans := 0
	for i := 0; i < n-1; i++ {
		for j := i + 1; j < n; j++ {
			if mask[i]&mask[j] == 0 {
				ans = max(ans, len(words[i])*len(words[j]))
			}
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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