4-1 Iterative Mergesort (9分)

How would you implement mergesort without using recursion?

The idea of iterative mergesort is to start from N sorted sublists of length 1, and each time to merge a pair of adjacent sublists until one sorted list is obtained. You are supposed to implement the key function of merging.

Format of functions:

void merge_pass( ElementType list[], ElementType sorted[], int N, int length );

The function merge_pass performs one pass of the merge sort that merges adjacent pairs of sublists from list into sorted. N is the number of elements in the list and length is the length of the sublists.

Sample program of judge:

#include <stdio.h>

#define ElementType int
#define MAXN 100

void merge_pass( ElementType list[], ElementType sorted[], int N, int length );

void output( ElementType list[], int N )
{
    int i;
    for (i=0; i<N; i++) printf("%d ", list[i]);
    printf("\n");
}

void  merge_sort( ElementType list[],  int N )
{
    ElementType extra[MAXN];  /* the extra space required */
    int  length = 1;  /* current length of sublist being merged */
    while( length < N ) { 
        merge_pass( list, extra, N, length ); /* merge list into extra */
        output( extra, N );
        length *= 2;
        merge_pass( extra, list, N, length ); /* merge extra back to list */
        output( list, N );
        length *= 2;
    }
} 


int main()
{
    int N, i;
    ElementType A[MAXN];

    scanf("%d", &N);
    for (i=0; i<N; i++) scanf("%d", &A[i]);
    merge_sort(A, N);
    output(A, N);

    return 0;
}

/* Your function will be put here */

Sample Input:

10
8 7 9 2 3 5 1 6 4 0

Sample Output:

7 8 2 9 3 5 1 6 0 4 
2 7 8 9 1 3 5 6 0 4 
1 2 3 5 6 7 8 9 0 4 
0 1 2 3 4 5 6 7 8 9 
0 1 2 3 4 5 6 7 8 9
void merge_pass( ElementType list[], ElementType sorted[], int N, int length )
{
	int now=0;
	int top=0;
	int left, right;
	while(now<N)
	{
		left=now;
		right=now+length;
		while(left<now+length && right<now+2*length && left<N && right<N)
		{
			if(list[left]<list[right])
			{
				sorted[top++]=list[left++];
			}
			else
			{
				sorted[top++]=list[right++];
			}
		}
		while(left<now+length && left<N)
		{
			sorted[top++]=list[left++];
		}
		while(right<now+2*length && right<N)
		{
			sorted[top++]=list[right++];
		}
		now=now+2*length;
	}
}
贵校的PTA作业，请自行绕道，以免不必要的麻烦