6-2 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

 

结尾无空行

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 

 

结尾无空行

解题思路：

这道题的思路跟归并排序的思路类似，不过归并排序用到递归，此题用到迭代的思想。

length是每次排序的时候需要分成length长度的块，比如说传入8 7 9 2 3 5 1 6 这个序列，如果给定的length为1，

那么需要将此序列分成10个块，如图：

设置a指向第i个block的头部，aa指向第i个block的尾部，b指向第i+length个block的头部，bb指向第i+length*2个block的尾部。

如果出现aa>N或者bb>N的情况，那么越界，将aa或者bb更改为N；

然后进入while循环，接下来的步骤便和归并排序类似，比list[a]和list[b]的大小，谁小就将它放入sorted[]中，直到a排完或者b排完。

i每次+=length*2，即每次比较两个块。

 

 

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