气泡排序逻辑，迭代次数

Question

I have the following code:

// C program for implementation of Bubble sort 
#include <stdio.h> 

void swap(int *xp, int *yp) 
{ 
    int temp = *xp; 
    *xp = *yp; 
    *yp = temp; 
} 

// A function to implement bubble sort 
void bubbleSort(int arr[], int n) 
{ 
   int i, j; 
   for (i = 0; i < n-1; i++)       

       // Last i elements are already in place    
       for (j = 0; j < n-i-1; j++)  
           if (arr[j] > arr[j+1]) 
              swap(&arr[j], &arr[j+1]); 
} 

/* Function to print an array */
void printArray(int arr[], int size) 
{ 
    int i; 
    for (i=0; i < size; i++) 
        printf("%d ", arr[i]); 
    printf("n"); 
} 

// Driver program to test above functions 
int main() 
{ 
    int arr[] = {64, 34, 25, 12, 22, 11, 90}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    bubbleSort(arr, n); 
    printf("Sorted array: \n"); 
    printArray(arr, n); 
    return 0; 
} 

The only part that makes me confused is where i < n-1 in the first for loop and J< ni-1 in the inner for loop inside the BubbleSort function. Why arent they both set to i <= n-1 and J<=ni-1 ? For instance, the first iteration would be a total of n= 7, therefore it means it should go through the loop for 6 times in the outer loop and 6 times in the inner for loop. But, without the <= sign it would only be 5 iterations per loop. On the website, it has illustrated that both loops do go through 6 iterations however Im not sure how would that happen without the <= in place.

Source: https://www.geeksforgeeks.org/bubble-sort/

Answer 1

Note the use of arr[j+1] . Let's say your array has n = 7 . Then when i = 0 and j = n - i - 1 = 6 , you would be accessing arr[j+1] = arr[6 + 1] = arr[7] . However, arr only had 7 elements to begin with, so index 7 is out of bounds since the indices begin at 0 , with arr[6] being the seventh element.

As for why it doesn't matter, the final element of the array is already swapped when comparing it to the next-to-last element. Or if the array only had 1 element, it is already sorted.