给定一个由小写字母组成的字符串。
游戏规则:
- 玩家可以选择一对相似的连续字符并删除它们。
- 有两个玩家在玩游戏,最后一步的玩家获胜。
任务是找到获胜者,如果 A 先走并且双方都发挥最佳。
例子:
Input: str = "kaak"
Output: B
Explanation:
Initial String: "kaak"
A's turn:
removes: "aa"
Remaining String: "kk"
B's turn:
removes: "kk"
Remaining String: ""
Since B was the last one to play
B is the winner.
Input: str = "kk"
Output: A
方法:我们可以使用堆栈来简化问题。
- 每次我们遇到一个与堆栈顶部的字符不同的字符,我们将其添加到堆栈中。
- 如果栈顶和下一个字符匹配,我们从栈中弹出字符并增加计数。
- 最后,我们只需要通过检查 count%2 来看看谁赢了。
下面是上述方法的实现:
C++
#include
using namespace std;
// Function to play the game
// and find the winner
void findWinner(string s)
{
int i, count = 0, n;
n = s.length();
stack st;
// ckecking the top of the stack with
// the i th character of the string
// add it to the stack if they are different
// otherwise increment count
for (i = 0; i < n; i++) {
if (st.empty() || st.top() != s[i]) {
st.push(s[i]);
}
else {
count++;
st.pop();
}
}
// Check who has won
if (count % 2 == 0) {
cout << "B" << endl;
}
else {
cout << "A" << endl;
}
}
// Driver code
int main()
{
string s = "kaak";
findWinner(s);
return 0;
}
Java
// Java implementation for above approach
import java.util.*;
class GFG
{
// Function to play the game
// and find the winner
static void findWinner(String s)
{
int i, count = 0, n;
n = s.length();
Stack st = new Stack();
// ckecking the top of the stack with
// the i th character of the string
// add it to the stack if they are different
// otherwise increment count
for (i = 0; i < n; i++)
{
if (st.isEmpty() ||
st.peek() != s.charAt(i))
{
st.push(s.charAt(i));
}
else
{
count++;
st.pop();
}
}
// Check who has won
if (count % 2 == 0)
{
System.out.println("B");
}
else
{
System.out.println("A");
}
}
// Driver code
public static void main(String[] args)
{
String s = "kaak";
findWinner(s);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach
# Function to play the game
# and find the winner
def findWinner(s) :
count = 0
n = len(s);
st = [];
# ckecking the top of the stack with
# the i th character of the string
# add it to the stack if they are different
# otherwise increment count
for i in range(n) :
if (len(st) == 0 or st[-1] != s[i]) :
st.append(s[i]);
else :
count += 1;
st.pop();
# Check who has won
if (count % 2 == 0) :
print("B");
else :
print("A");
# Driver code
if __name__ == "__main__" :
s = "kaak";
findWinner(s);
# This code is contributed by AnkitRai01
C#
// C# implementation for above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to play the game
// and find the winner
static void findWinner(String s)
{
int i, count = 0, n;
n = s.Length;
Stack st = new Stack();
// ckecking the top of the stack with
// the i th character of the string
// add it to the stack if they are different
// otherwise increment count
for (i = 0; i < n; i++)
{
if (st.Count == 0 ||
st.Peek() != s[i])
{
st.Push(s[i]);
}
else
{
count++;
st.Pop();
}
}
// Check who has won
if (count % 2 == 0)
{
Console.WriteLine("B");
}
else
{
Console.WriteLine("A");
}
}
// Driver code
public static void Main(String[] args)
{
String s = "kaak";
findWinner(s);
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
B
如果您希望与专家一起参加现场课程,请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。