Given an array that is initially stored in one stack, sort it with one additional stacks (total 2 stacks).

After sorting the original stack should contain the sorted integers and from top to bottom the integers are sorted in ascending order.

Assumptions:

• The given stack is not null.

Requirements:

• No additional memory, time complexity = O(n ^ 2).

基於selection sort，一個stack即當buffer又當output，每次倒的時候記錄global min及出現次數（以防重複元素）

time: O(n^2), space: O(n)

public class Solution {
  public void sort(LinkedList<Integer> s1) {
    LinkedList<Integer> s2 = new LinkedList<Integer>();
    
 
// Write your solution here.
    if(s1 == null || s1.size() < 0) {
      return;
    }
    int cnt = 0;
    while(!s1.isEmpty()) {
      int globalMin = Integer.MAX_VALUE;
      while(!s1.isEmpty()) {
        int tmp = s1.pop();
        if(tmp < globalMin) {
          globalMin = tmp;
        }
        s2.push(tmp);
      }
      
 
while(!s2.isEmpty() && s2.peek() >= globalMin) {
        int tmp = s2.pop();
        if(tmp == globalMin) {
          cnt++;
        } else {
          s1.push(tmp);
        }
      }
      while(cnt > 0) {
        s2.push(globalMin);
        cnt--;
      }
    }
    
    while(!s2.isEmpty()) {
      s1.push(s2.pop());
    }
  }
}