【leetcode】318. Maximum Product of Word Lengths

2024年5月22日 294次阅读

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.

根据题目的描述，我们可知，最耗时的地方在于判断两个字符串中是否存在重复的字符。

一提到判断重复字符，脑海中马上浮现哈希大法。

又看到题目要保证字符串均为小写字母，也就说总共26个字母，我们需要26位，一个int足矣。

我们自定义，如果某字符出现，该bit置1，否则置0.

则比较两个字符串是否有重复字符，就是让这两个标志int 进行与操作，结果为0则表明没有重复字符。

好的，本题的难点已结束。code如下：

易知时间复杂度为O(n2)，空间复杂度为O(n)，其中n为所有字符串的个数。

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int isDuplicate[2000] = {0};
        int nSize = words.size();
        for(int i = 0; i < nSize; i++)
        {
            int length = words[i].length();
            for(int j = 0; j < length; j++)
                isDuplicate[i] |= 1 << (words[i][j] - 'a');
        }
        int maxMul = 0, tempMul = 0;
        for(int i = 0; i < nSize -1; i++)
            for(int j = i +1; j < nSize; j++)
                if((isDuplicate[i] & isDuplicate[j]) == 0)
                {
                    tempMul = words[i].length()*words[j].length();
                    if(maxMul < tempMul)
                        maxMul = tempMul;
                }
        return maxMul;
    }
};

Submission Details

174 / 174 test cases passed.

Status:
Accepted

Runtime: 104 ms

Submitted: 0 minutes ago

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