4-1 Iterative Mergesort (9分)

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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 fromlist into sorted. N is the number of elements in thelist and length is the length of the sublists.

Sample program of judge:

#include <stdio.h>#define ElementType int#define MAXN 100void 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:

108 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 temp;if(length>N)return ;int i,j=0,count1=0,count2=0;int M=N;count1=0;count2=length;int start,end;int t1,t2;//printf("%d %d %d\n",count2,length,N);for(i=0;i<(N-1)/(2*length)+1;i++){t1=0;t2=0;if(i!=(N-1)/(2*length)){start=i*length*2;end=start+length;while(t1+t2!=2*length){if((list[t1+start]<=list[t2+end]||t2==length)&&t1<length){sorted[j++]=list[start+t1++];}if((list[t1+start]>=list[t2+end]||t1==length)&&t2<length){sorted[j++]=list[end+t2++];}}}else{if((N-1)%(2*length)+1<=length){while(j++<N)sorted[j-1]=list[j-1];}else{start=N-(N-1)%(2*length)-1;end=start+length;while(t1+t2!=N-start){if((list[t1+start]<=list[t2+end]||t2==N-end)&&t1<length){sorted[j++]=list[start+t1++];}if((list[t1+start]>=list[t2+end]||t1==length)&&t2<N-end){sorted[j++]=list[end+t2++];}}}}}}

感想：最重要的一点，理解题意，就算最后剩余的个数不足以完成归并排序，也要强行归并起来！！！不然最后一个点过不了

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