为什么我的替代排序算法不起作用？

Question

I was trying to implement a sorting code so i tried something completely without looking at any other reference codes or anything. Please tell me why my code doesnt show any output? It just runs for sometime and then stops suddenly.

I want to know where i am going wrong. I wanted to take any array as an input and sort the numbers in it in ascending order. So i just iterated throughput the array and compared adjacent elements and swaped them. Then i tried printing the numbers using their indices but it did not print anything on the screen.

#include <iostream>
using namespace std;
int main() {
  int arr[6]={1,23,2,32,4,12};
  int i=0;
  while(i<6)
  {
    if(arr[i]>arr[i+1])
    {
      int x = arr[i];
      arr[i]=arr[i+1];
      arr[i+1]=x;
    }
    else
      continue;
    i++;
  }
  cout<<arr[0];
  cout<<arr[1];
  cout<<arr[2];
  cout<<arr[3];
  cout<<arr[4];
  cout<<arr[5];
  return 0;
}

I expected that the numbers will be printed in ascending order but nothing happened. And also tell me how to print an array all at once. Thank you

Answer 1

Your sorting algorithm doesn't work because it performs only one pass of a variant of Bubble Sort or Insertion Sort. There has to be one more loop wrapped around your loop to repeat the operation N-1 times.

There are a number of ways to explain why multiple passes are required. Here is one explanation. Whenever we perform this step:

if (arr[i] > arr[i+1])
{
  int x = arr[i];
  arr[i] = arr[i+1];
  arr[i+1] = x;
}

we transfer the arr[i+1] element into the arr[i] position. By doing so, we effectively skip the arr[i+1] element.

Here is what I mean. Suppose arr[i] is called x , arr[i+1] is called y and arr[i+2] is called z . We start with xyz , and exchange x and y to make yxz . But then on the next iteration of the loop, we compare x and z and have forgotten about y ; x has moved forward in the array, pushing down y . Why is that a problem? Because y and z have not been compared; and those two are not necessarily in sorted order!

Suppose we have { 3, 2, 0 }. We compare 3 and 2, and swap them, giving us { 2, 3, 0}. Then we move on to comparing 3 and 0: we swap those and get { 2, 0, 3 }. See the problem? We neglected to deal with 2 and 0. That requires another pass through the array.

There is a family of algorithms which work by repeatedly scanning through an array and exchanging items. I suggest studying the following of the common algorithms in this family: insertion sort , selection sort and Shell sort. Also look at bubble sort, in order to understand why it's a poor algorithm compared to either insertion or selection sort that it is closely related to.

Answer 2

1. your loop condition is incorrect - the last iteration will compare arr[5] and arr[6] which gives index out of bound.

2. you update your iterator "i" only if you swap, so if your loop encounters the case where arr[i] <= arr[i+1], your iterator will never get updated and your program will run infinitely.

3. your algorithm implements just a first iteration of bubble sort ie it bubbles up (from left to right) the largest number and stops. so the final array looks like [1, 2, 23, 4, 12, 32]