6-2 Iterative Mergesort （25 分）-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_40215528/article/details/84996508

本文介绍了一种不使用递归的归并排序算法实现方法，通过迭代的方式将多个已排序的子列表合并成一个完全排序的列表。文章详细解释了算法的工作原理，并提供了一个具体的函数merge_pass，用于执行归并排序的合并过程。

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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 listand 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 ){
    for (int i=0; i<N; i=i+2*length) {
        int p1,p2,p3;
        p1=i;
        p2=i+length;
        p3=p1;
        while (p1<i+length&&p2<i+length+length&&p1<N&&p2<N) {
            if(list[p1]<list[p2]){
                sorted[p3]=list[p1];
                p3++;
                p1++;
            }
            else{
                sorted[p3]=list[p2];
                p3++;
                p2++;
            }
        }
        if(p1==i+length){
            while (p2<i+length+length) {
                sorted[p3]=list[p2];
                p3++;
                p2++;
            }
        }
        if(p2==i+length+length){
            while (p1<i+length) {
                sorted[p3]=list[p1];
                p3++;
                p1++;
            }
        }
        if(p2>=N){
            while (p1<i+length) {
                sorted[p3]=list[p1];
                p3++;
                p1++;
            }
        }
        if(p1>=N){
            while (p2<i+length) {
                sorted[p3]=list[p2];
                p3++;
                p2++;
            }
        }
    }
}