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105 lines
2.4 KiB
Markdown
105 lines
2.4 KiB
Markdown
# [914. X of a Kind in a Deck of Cards](https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/)
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## 题目
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In a deck of cards, each card has an integer written on it.
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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:
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- Each group has exactly `X` cards.
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- All the cards in each group have the same integer.
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**Example 1**:
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```
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Input: deck = [1,2,3,4,4,3,2,1]
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Output: true
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Explanation: Possible partition [1,1],[2,2],[3,3],[4,4].
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```
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**Example 2**:
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```
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Input: deck = [1,1,1,2,2,2,3,3]
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Output: false´
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Explanation: No possible partition.
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```
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**Example 3**:
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```
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Input: deck = [1]
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Output: false
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Explanation: No possible partition.
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```
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**Example 4**:
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```
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Input: deck = [1,1]
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Output: true
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Explanation: Possible partition [1,1].
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```
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**Example 5**:
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```
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Input: deck = [1,1,2,2,2,2]
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Output: true
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Explanation: Possible partition [1,1],[2,2],[2,2].
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```
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**Constraints**:
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- `1 <= deck.length <= 10^4`
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- `0 <= deck[i] < 10^4`
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## 题目大意
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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：
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- 每组都有 X 张牌。
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- 组内所有的牌上都写着相同的整数。
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仅当你可选的 X >= 2 时返回 true。
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## 解题思路
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- 给定一副牌，要求选出数字 X，使得每组都有 X 张牌，每组牌的数字都相同。当 X ≥ 2 的时候，输出 true。
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- 通过分析题目，我们可以知道，只有当 X 为所有 count 的约数，即所有 count 的最大公约数的约数时，才存在可能的分组。因此我们只要求出所有 count 的最大公约数 g，判断 g 是否大于等于 2 即可，如果大于等于 2，则满足条件，否则不满足。
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- 时间复杂度：O(NlogC)，其中 N 是卡牌的个数，C 是数组 deck 中数的范围，在本题中 C 的值为 10000。求两个数最大公约数的复杂度是 O(logC)，需要求最多 N - 1 次。空间复杂度：O(N + C) 或 O(N)。
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## 代码
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```go
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package leetcode
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func hasGroupsSizeX(deck []int) bool {
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if len(deck) < 2 {
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return false
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}
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m, g := map[int]int{}, -1
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for _, d := range deck {
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m[d]++
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}
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for _, v := range m {
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if g == -1 {
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g = v
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} else {
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g = gcd(g, v)
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}
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}
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return g >= 2
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}
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func gcd(a, b int) int {
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if a == 0 {
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return b
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}
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return gcd(b%a, a)
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}
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``` |