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feat: add solutions to lc problem: No.2974 (#2149)
No.2974.Minimum Number Game
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‎solution/2900-2999/2974.Minimum Number Game/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：模拟 + 优先队列（小根堆）**
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我们可以将数组 $nums$ 中的元素依次放入一个小根堆中，每次从小根堆中取出两个元素 $a$ 和 $b$，然后依次将 $b$ 和 $a$ 放入答案数组中，直到小根堆为空。
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时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def numberGame(self, nums: List[int]) -> List[int]:
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heapify(nums)
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ans = []
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while nums:
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a, b = heappop(nums), heappop(nums)
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ans.append(b)
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ans.append(a)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[] numberGame(int[] nums) {
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PriorityQueue<Integer> pq = new PriorityQueue<>();
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for (int x : nums) {
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pq.offer(x);
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}
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int[] ans = new int[nums.length];
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int i = 0;
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while (!pq.isEmpty()) {
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int a = pq.poll();
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ans[i++] = pq.poll();
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ans[i++] = a;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> numberGame(vector<int>& nums) {
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priority_queue<int, vector<int>, greater<int>> pq;
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for (int x : nums) {
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pq.push(x);
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}
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vector<int> ans;
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while (pq.size()) {
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int a = pq.top();
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pq.pop();
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int b = pq.top();
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pq.pop();
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ans.push_back(b);
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ans.push_back(a);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func numberGame(nums []int) (ans []int) {
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pq := &hp{nums}
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heap.Init(pq)
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for pq.Len() > 0 {
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a := heap.Pop(pq).(int)
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b := heap.Pop(pq).(int)
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ans = append(ans, b)
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ans = append(ans, a)
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}
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return
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}
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type hp struct{ sort.IntSlice }
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func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
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func (h *hp) Pop() interface{} {
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old := h.IntSlice
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n := len(old)
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x := old[n-1]
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h.IntSlice = old[0 : n-1]
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return x
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}
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func (h *hp) Push(x interface{}) {
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h.IntSlice = append(h.IntSlice, x.(int))
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}
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```
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### **TypeScript**
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```ts
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function numberGame(nums: number[]): number[] {
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const pq = new MinPriorityQueue();
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for (const x of nums) {
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pq.enqueue(x);
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}
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const ans: number[] = [];
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while (pq.size()) {
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const a = pq.dequeue().element;
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const b = pq.dequeue().element;
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ans.push(b, a);
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2974.Minimum Number Game/README_EN.md‎

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## Solutions
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**Solution 1: Simulation + Priority Queue (Min Heap)**
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We can put the elements in the array $nums$ into a min heap one by one, and each time take out two elements $a$ and $b$ from the min heap, then put $b$ and $a$ into the answer array in turn, until the min heap is empty.
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Time complexity is $O(n \times \log n)$, and space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def numberGame(self, nums: List[int]) -> List[int]:
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heapify(nums)
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ans = []
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while nums:
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a, b = heappop(nums), heappop(nums)
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ans.append(b)
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ans.append(a)
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int[] numberGame(int[] nums) {
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PriorityQueue<Integer> pq = new PriorityQueue<>();
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for (int x : nums) {
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pq.offer(x);
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}
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int[] ans = new int[nums.length];
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int i = 0;
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while (!pq.isEmpty()) {
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int a = pq.poll();
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ans[i++] = pq.poll();
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ans[i++] = a;
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> numberGame(vector<int>& nums) {
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priority_queue<int, vector<int>, greater<int>> pq;
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for (int x : nums) {
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pq.push(x);
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}
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vector<int> ans;
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while (pq.size()) {
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int a = pq.top();
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pq.pop();
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int b = pq.top();
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pq.pop();
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ans.push_back(b);
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ans.push_back(a);
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func numberGame(nums []int) (ans []int) {
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pq := &hp{nums}
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heap.Init(pq)
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for pq.Len() > 0 {
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a := heap.Pop(pq).(int)
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b := heap.Pop(pq).(int)
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ans = append(ans, b)
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ans = append(ans, a)
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}
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return
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}
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type hp struct{ sort.IntSlice }
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func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
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func (h *hp) Pop() interface{} {
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old := h.IntSlice
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n := len(old)
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x := old[n-1]
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h.IntSlice = old[0 : n-1]
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return x
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}
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func (h *hp) Push(x interface{}) {
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h.IntSlice = append(h.IntSlice, x.(int))
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}
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```
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### **TypeScript**
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```ts
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function numberGame(nums: number[]): number[] {
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const pq = new MinPriorityQueue();
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for (const x of nums) {
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pq.enqueue(x);
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}
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const ans: number[] = [];
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while (pq.size()) {
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const a = pq.dequeue().element;
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const b = pq.dequeue().element;
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ans.push(b, a);
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2974.Minimum Number Game/Solution.cpp‎

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class Solution {
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public:
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vector<int> numberGame(vector<int>& nums) {
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priority_queue<int, vector<int>, greater<int>> pq;
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for (int x : nums) {
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pq.push(x);
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}
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vector<int> ans;
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while (pq.size()) {
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int a = pq.top();
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pq.pop();
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int b = pq.top();
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pq.pop();
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ans.push_back(b);
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ans.push_back(a);
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}
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return ans;
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}
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};

‎solution/2900-2999/2974.Minimum Number Game/Solution.go‎

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func numberGame(nums []int) (ans []int) {
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pq := &hp{nums}
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heap.Init(pq)
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for pq.Len() > 0 {
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a := heap.Pop(pq).(int)
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b := heap.Pop(pq).(int)
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ans = append(ans, b)
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ans = append(ans, a)
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}
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return
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}
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type hp struct{ sort.IntSlice }
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func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
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func (h *hp) Pop() interface{} {
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old := h.IntSlice
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n := len(old)
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x := old[n-1]
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h.IntSlice = old[0 : n-1]
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return x
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}
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func (h *hp) Push(x interface{}) {
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h.IntSlice = append(h.IntSlice, x.(int))
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}

‎solution/2900-2999/2974.Minimum Number Game/Solution.java‎

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class Solution {
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public int[] numberGame(int[] nums) {
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PriorityQueue<Integer> pq = new PriorityQueue<>();
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for (int x : nums) {
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pq.offer(x);
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}
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int[] ans = new int[nums.length];
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int i = 0;
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while (!pq.isEmpty()) {
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int a = pq.poll();
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ans[i++] = pq.poll();
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ans[i++] = a;
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}
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return ans;
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}
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}

‎solution/2900-2999/2974.Minimum Number Game/Solution.py‎

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class Solution:
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def numberGame(self, nums: List[int]) -> List[int]:
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heapify(nums)
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ans = []
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while nums:
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a, b = heappop(nums), heappop(nums)
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ans.append(b)
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ans.append(a)
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return ans

‎solution/2900-2999/2974.Minimum Number Game/Solution.ts‎

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function numberGame(nums: number[]): number[] {
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const pq = new MinPriorityQueue();
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for (const x of nums) {
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pq.enqueue(x);
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}
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const ans: number[] = [];
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while (pq.size()) {
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const a = pq.dequeue().element;
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const b = pq.dequeue().element;
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ans.push(b, a);
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}
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return ans;
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}

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