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Commit c93f25e

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feat: add solutions to lc problem: No.0292.Nim Game
1 parent cdfac37 commit c93f25e

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6 files changed

+75
-4
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6 files changed

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‎solution/0200-0299/0292.Nim Game/README.md‎

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<li><code>1 <= n <= 2<sup>31</sup> - 1</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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第一个得到 4 的倍数(即 n % 4 == 0)的将会输掉比赛。
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证明:
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1. 当 `n == 4`,无论第一个玩家选择 1/2/3 哪个数字,第二个玩家总能选择剩下的数字,**第一个玩家将会输掉比赛**。
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1. 当 `4 < n < 8`,即 (n = 5,6,7),第一个玩家可以相应地将数字减少为 4,那么 4 这个死亡数字给到了第二个玩家,第二个玩家将会输掉比赛。
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1. 当 `n == 8`,无论第一个玩家选择 1/2/3 哪个数字,都会把 `4 < n < 8` 的数字留给第二个,**第一个玩家将会输掉比赛**。
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1. ...
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1. 依次类推,当玩家拿到 n 这个数字,且 n 能被 4 整除,即 `n % 4 == 0`,他将会输掉比赛,否则他将赢得比赛。
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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 canWinNim(self, n: int) -> bool:
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return n % 4 != 0
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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 boolean canWinNim(int n) {
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return n % 4 != 0;
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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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bool canWinNim(int n) {
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return n % 4 != 0;
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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 canWinNim(n int) bool {
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return n%4 != 0
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}
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```
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### **...**

‎solution/0200-0299/0292.Nim Game/README_EN.md‎

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### **Python3**
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```python
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class Solution:
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def canWinNim(self, n: int) -> bool:
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return n % 4 != 0
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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 boolean canWinNim(int n) {
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return n % 4 != 0;
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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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bool canWinNim(int n) {
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return n % 4 != 0;
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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 canWinNim(n int) bool {
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return n%4 != 0
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}
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```
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### **...**
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class Solution {
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public:
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bool canWinNim(int n) {
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return n % 4 != 0;
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}
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};
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func canWinNim(n int) bool {
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return n%4 != 0
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}
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class Solution {
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public boolean canWinNim(int n) {
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return (n & 3) != 0;// n%4 != 0
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return n % 4 != 0;
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}
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}
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class Solution:
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def canWinNim(self, n: int) -> bool:
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return n % 4 != 0

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