给定一个整数堆栈和一个整数K ，任务是使用另一个堆栈按给定堆栈的元素对K取模的递增顺序对它们进行排序。如果两个数字的余数相同，则较小的数字应排在第一位。
例子
Input: stack = {10, 3, 2, 6, 12}, K = 4
Output: 12 2 6 10 3
{12, 2, 6, 10, 3} is the required sorted order as the modulo
of these elements with K = 4 is {2, 3, 2, 2, 0}
Input: stack = {3, 4, 5, 10, 11, 1}, K = 3
Output: 3 1 4 10 5 11
方法：本文讨论了一种使用另一个临时堆栈对堆栈元素进行排序的方法，此处可以使用相同的方法根据元素对K的模数对元素进行排序，唯一的区别是当比较元素时给出相同的模值，然后将基于它们的值进行比较。
下面是上述方法的实现：
C++
// C++ implementation of the
// above approach
#include
using namespace std;
// Function to sort the stack using
// another stack based on the
// values of elements modulo k
void sortStack(stack& input, int k)
{
stack tmpStack;
while (!input.empty()) {
// Pop out the first element
int tmp = input.top();
input.pop();
// While temporary stack is not empty
while (!tmpStack.empty()) {
int tmpStackMod = tmpStack.top() % k;
int tmpMod = tmp % k;
// The top of the stack modulo k is
// greater than (temp & k) or if they
// are equal then compare the values
if ((tmpStackMod > tmpMod)
|| (tmpStackMod == tmpMod
&& tmpStack.top() > tmp)) {
// Pop from temporary stack and push
// it to the input stack
input.push(tmpStack.top());
tmpStack.pop();
}
else
break;
}
// Push temp in tempory of stack
tmpStack.push(tmp);
}
// Push all the elements in the original
// stack to get the ascending order
while (!tmpStack.empty()) {
input.push(tmpStack.top());
tmpStack.pop();
}
// Print the sorted elements
while (!input.empty()) {
cout << input.top() << " ";
input.pop();
}
}
// Driver code
int main()
{
stack input;
input.push(10);
input.push(3);
input.push(2);
input.push(6);
input.push(12);
int k = 4;
sortStack(input, k);
return 0;
} Java
// Java implementation of the
// above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to sort the stack using
// another stack based on the
// values of elements modulo k
static void sortStack(Stack input, int k)
{
Stack tmpStack = new Stack();
while (!input.isEmpty())
{
// Pop out the first element
int tmp = input.peek();
input.pop();
// While temporary stack is not empty
while (!tmpStack.isEmpty())
{
int tmpStackMod = tmpStack.peek() % k;
int tmpMod = tmp % k;
// The top of the stack modulo k is
// greater than (temp & k) or if they
// are equal then compare the values
if ((tmpStackMod > tmpMod) ||
(tmpStackMod == tmpMod &&
tmpStack.peek() > tmp))
{
// Pop from temporary stack and push
// it to the input stack
input.push(tmpStack.peek());
tmpStack.pop();
}
else
break;
}
// Push temp in tempory of stack
tmpStack.push(tmp);
}
// Push all the elements in the original
// stack to get the ascending order
while (!tmpStack.isEmpty())
{
input.push(tmpStack.peek());
tmpStack.pop();
}
// Print the sorted elements
while (!input.empty())
{
System.out.print(input.peek() + " ");
input.pop();
}
}
// Driver code
public static void main(String args[])
{
Stack input = new Stack();
input.push(10);
input.push(3);
input.push(2);
input.push(6);
input.push(12);
int k = 4;
sortStack(input, k);
}
}
// This code is contributed by adityapande88 Python3
# Python3 implementation of the
# above approach
# Function to sort the stack using
# another stack based on the
# values of elements modulo k
def sortStack(input1, k):
tmpStack = []
while (len(input1) != 0):
# Pop out the first element
tmp = input1[-1]
input1.pop()
# While temporary stack is
# not empty
while (len(tmpStack) != 0):
tmpStackMod = tmpStack[-1] % k
tmpMod = tmp % k
# The top of the stack modulo
# k is greater than (temp & k)
# or if they are equal then
# compare the values
if ((tmpStackMod > tmpMod) or
(tmpStackMod == tmpMod and
tmpStack[-1] > tmp)):
# Pop from temporary stack
# and push it to the input
# stack
input1.append(tmpStack[-1])
tmpStack.pop()
else:
break
# Push temp in tempory of stack
tmpStack.append(tmp)
# Push all the elements in
# the original stack to get
# the ascending order
while (len(tmpStack) != 0):
input1.append(tmpStack[-1])
tmpStack.pop()
# Print the sorted elements
while (len(input1) != 0):
print(input1[-1], end = " ")
input1.pop()
# Driver code
if __name__ == "__main__":
input1 = []
input1.append(10)
input1.append(3)
input1.append(2)
input1.append(6)
input1.append(12)
k = 4
sortStack(input1, k)
# This code is contributed by ChitranayalC#
// C# implementation of the
// above approach
using System;
using System.Collections;
class GFG{
// Function to sort the stack using
// another stack based on the
// values of elements modulo k
static void sortStack(Stack input,
int k)
{
Stack tmpStack = new Stack();
while(input.Count != 0)
{
// Pop out the first element
int tmp = (int)input.Peek();
input.Pop();
// While temporary stack is not empty
while (tmpStack.Count != 0)
{
int tmpStackMod = (int)tmpStack.Peek() % k;
int tmpMod = tmp % k;
// The top of the stack modulo k is
// greater than (temp & k) or if they
// are equal then compare the values
if ((tmpStackMod > tmpMod) ||
(tmpStackMod == tmpMod &&
(int)tmpStack.Peek() > tmp))
{
// Pop from temporary stack and push
// it to the input stack
input.Push((int)tmpStack.Peek());
tmpStack.Pop();
}
else
break;
}
// Push temp in tempory of stack
tmpStack.Push(tmp);
}
// Push all the elements in the original
// stack to get the ascending order
while (tmpStack.Count != 0)
{
input.Push((int)tmpStack.Peek());
tmpStack.Pop();
}
// Print the sorted elements
while (input.Count != 0)
{
Console.Write((int)input.Peek() + " ");
input.Pop();
}
}
// Driver Code
public static void Main(string[] args)
{
Stack input = new Stack();
input.Push(10);
input.Push(3);
input.Push(2);
input.Push(6);
input.Push(12);
int k = 4;
sortStack(input, k);
}
}
// This code is contributed by rutvik_56输出：
12 2 6 10 3