914 X of a Kind in a Deck of Cards 卡牌分组
Description:
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:
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
题目描述:
给定一副牌，每张牌上都写着一个整数。
此时，你需要选定一个数字 X，使我们可以将整副牌按下述规则分成 1 组或更多组：
每组都有 X 张牌。
组内所有的牌上都写着相同的整数。
仅当你可选的 X >= 2 时返回 true。
示例 :
示例 1：
输入：[1,2,3,4,4,3,2,1]
输出：true
解释：可行的分组是 [1,1]，[2,2]，[3,3]，[4,4]
示例 2：
输入：[1,1,1,2,2,2,3,3]
输出：false
解释：没有满足要求的分组。
示例 3：
输入：[1]
输出：false
解释：没有满足要求的分组。
示例 4：
输入：[1,1]
输出：true
解释：可行的分组是 [1,1]
示例 5：
输入：[1,1,2,2,2,2]
输出：true
解释：可行的分组是 [1,1]，[2,2]，[2,2]
提示：
1 <= deck.length <= 10000
0 <= deck[i] < 10000
思路:
用 map(Counter)记录下每个元素的出现次数, 取所有出现次数的最小公倍数, 返回最小公倍数是否大于等于 2即可
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
bool hasGroupsSizeX(vector<int>& deck)
{
map<int, int> m;
for (auto num : deck) m[num]++;
int result = m[deck[0]];
for (auto item : m)
{
result = gcd(result, item.second);
if (result == 1) return false;
}
return true;
}
private:
int gcd(int a, int b)
{
return a > b ? gcd(b, a) : (a == 0 ? b : gcd(b % a, a));
}
};
Java:
class Solution {
public boolean hasGroupsSizeX(int[] deck) {
Map<Integer, Integer> map = new HashMap<>();
for (int num : deck) map.put(num, map.getOrDefault(num, 0) + 1);
int result = map.get(deck[0]);
for (int value : map.values()) result = gcd(result, value);
return result > 1;
}
private int gcd(int a, int b) {
return a > b ? gcd(b, a) : (a == 0 ? b : gcd(b % a, a));
}
}
Python:
class Solution:
def hasGroupsSizeX(self, deck: List[int]) -> bool:
c = collections.Counter(deck)
return any((all((d % i == 0 for d in c.values())) for i in range(2, 5000)))
