6-15 Iterative Mergesort (25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 

代码：

void merge_pass( ElementType list[], ElementType sorted[], int N, int length ) {
    for(int i = 0;i < N;i += length * 2) {
        int a = i,b = i + length,j = 0;
        int aa = i + length,bb = i + length * 2;
        if(aa > N) aa = N;
        if(bb > N) bb = N;
        while(a < aa && b < bb) {
            if(list[a] < list[b]) {
                sorted[i + j ++] = list[a ++];
            }
            else sorted[i + j ++] = list[b ++];
        }
        while(a < aa) {
            sorted[i + j ++] = list[a ++];
        }
        while(b < bb) {
            sorted[i + j ++] = list[b ++];
        }
    }
}

原文地址：https://www.cnblogs.com/8023spz/p/12255180.html