914. X of a Kind in a Deck of Cards

In a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

  • Each group has exactly X cards.
  • All the cards in each group have the same integer.

Example 1:

Input: [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4]

Example 2:

Input: [1,1,1,2,2,2,3,3]
Output: false Explanation: No possible partition.

Example 3:

Input: [1]
Output: false Explanation: No possible partition.

Example 4:

Input: [1,1]
Output: true Explanation: Possible partition [1,1]

Example 5:

Input: [1,1,2,2,2,2]
Output: true Explanation: Possible partition [1,1],[2,2],[2,2]

Note:

  1. 1 <= deck.length <= 10000

  2. 0 <= deck[i] < 10000

    这道题给了一堆牌,问我们能不能将这副牌分成若干堆,每堆均有X个,且每堆的牌数字都相同(这里不考虑花色)。既然要将相同的牌归类,肯定要统计每种牌出现的个数,所以使用一个 HashMap 来建立牌跟其出现次数之间的映射。由于每堆X个,则若果某张牌的个数小于X,则肯定无法分,所以X的范围是可以确定的,为 [2, mn],其中 mn 是数量最少的牌的个数。遍历一遍 HashMap,找出最小的映射值 mn,若 mn 小于2,可以直接返回 false。否则就从2遍历到 mn,依次来检验候选值X。检验的方法是看其他每种牌的个数是否能整除候选值X,不一定非要相等,比如 [1, 1, 2, 2, 2, 2], K=2 时就可以分为三堆 [1, 1], [2, 2], [2, 2],即相同的牌也可以分到其他堆里,所以只要每种牌的个数能整除X即可,一旦有牌数不能整除X了,则当前X一定不行,还得继续检验下一个X值;若所有牌数都能整除X,可以返回 true。循环结束后返回 false,参见代码如下:
    解法一:

    class Solution {
    public:
    bool hasGroupsSizeX(vector& deck) {

       unordered_map<int, int> cardCnt;
       for (int card : deck) ++cardCnt[card];
       int mn = INT_MAX;
       for (auto &a : cardCnt) mn = min(mn, a.second);
       if (mn < 2) return false;
       for (int i = 2; i <= mn; ++i) {
           bool success = true;
           for (auto &a : cardCnt) {
               if (a.second % i != 0) {
                   success = false;
                   break;
               }
           }
           if (success) return true;
       }
       return false;
    

    }
    };

    上面的解法是博主自己的解法,论坛上好多人使用了一个基于最大公约数 Greatest Common Divisor 的解法,写起来很简洁,但需要记住最大公约函数的写法,或者直接使用内置的 gcd 函数(感觉有点耍赖哈~)。其实原理都差不多,这里是找每种牌数之间的最大公约数,只要这个 gcd 是大于1的,就表示可以找到符合题意的X,参见代码如下:
    解法二:

    class Solution {
    public:
    bool hasGroupsSizeX(vector& deck) {

       unordered_map<int, int> cardCnt;
       for (int card : deck) ++cardCnt[card];
       int res = 0;
       for (auto &a : cardCnt) {
           res = gcd(a.second, res);
       }
       return res > 1;
    

    }
    int gcd(int a, int b) {

       return a == 0 ? b : gcd(b % a, a);
    

    }
    };

    Github 同步地址:

https://github.com/grandyang/leetcode/issues/914

参考资料:

https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/

https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/discuss/175845/C%2B%2BJavaPython-Greatest-Common-Divisor

https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/discuss/309992/Java-Easy-to-understand-2-ms-faster-than-98.92-38.5-MB-less-than-99.23

LeetCode All in One 题目讲解汇总(持续更新中…)


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