[LeetCode] 914. X of a Kind in a Deck of Cards
grandyang opened this issue · 0 comments
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 <= deck.length <= 10000
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<int>& 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<int>& 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:
参考资料:
https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/