Sort With 3 Stacks - Medium

三栈排序算法

最新推荐文章于 2025-08-08 16:32:18 发布

转载最新推荐文章于 2025-08-08 16:32:18 发布 · 128 阅读

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原文链接：http://www.cnblogs.com/fatttcat/p/10216064.html

本文介绍了一种使用三个栈进行排序的算法，通过不断倒换元素并寻找全局最小值实现排序，时间复杂度为O(n^2)。同时，还探讨了使用归并排序改进算法，达到O(nlogn)的时间复杂度。

Given one stack with integers, sort it with two additional stacks (total 3 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(nlog(n)).

M1: 普通作法，来回倒，s2作为buffer，s3作为output，最后再把s3中排完序的数字倒入s1

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

public class Solution {
  public void sort(LinkedList<Integer> s1) {
    LinkedList<Integer> s2 = new LinkedList<Integer>();
    LinkedList<Integer> s3 = new LinkedList<Integer>();
    // Write your solution here.
    if(s1 == null) {
      return;
    }
    
    while(!s1.isEmpty() || !s2.isEmpty()) {
      int globalMin = Integer.MAX_VALUE;
      while(!s1.isEmpty()) {
        int tmp = s1.pop();
        globalMin = tmp < globalMin ? tmp : globalMin;
        s2.push(tmp);
      }
      while(!s2.isEmpty()) {
        int tmp = s2.pop();
        if(tmp != globalMin) {
          s1.push(tmp);
        } else {
          s3.push(globalMin);
        }
      }
    }
    
    while(!s3.isEmpty()) {
      s1.push(s3.pop());
    }
  }
}

M2: merge sort

time: O(nlogn), space: O()

转载于:https://www.cnblogs.com/fatttcat/p/10216064.html