二路归并算法非递归C实现

算法思想是Horowitz E.Sahni S. Fundamentals of Data Structures, 1976上的。二路归并的递归实现相对简单，但是非递归实现却有些绕脑。所以尝试写代码并做一些跟踪，才最终理解了这个算法。  

#include <stdio.h>

#include <stdlib.h>

// X[0] is guard

void Print(int *X, int n)

{

printf("/n");

for(int i=1; i<=n; i++){

printf(" %d", X[i]);

}

}

// merge X[l]..X[m], X[m+1]..x
to Z

void MERGE(int* X

, int l/*first-sequence-start-postion, 1-based*/

, int m/*first-sequence-length, 1-based*/

, int n/*last-sequence-position, 1-based*/

, int* Z)

{

printf("/nMERGE(%08X, %d, %d, %d, %08X)", X, l, m, n, Z);

int i, j, k, t;

for(i=l, j=m+1, k=l; i<=m && j<=n; ){

if(X[i]<=X[j]){Z[k++]=X[i]; ++i;}

else{Z[k++]=X[j]; ++j;}

}

if(i>m){

printf("/nMerge: i=%d > m=%d", i, m);

for(t=j; t<=n; ++t)

Z[k++] = X[t];

}else{

printf("/nMerge: i=%d <= m=%d", i, m);

for(t=i; t<=m; ++t)

Z[k++] = X[t];

}

}

// One pass for sort X[1+l*k]..x[1+l-1+l*k]

void MPASS(int* X, int* Y, int n, int l)

{

printf("/nMPASS(%08X, %08X, %d, %d)", X, Y, n, l);

int i=1;

printf("/nMPASS(): i=%d", i);

while(i<=n-2*l+1){

MERGE(X, i, i+l-1, i+2*l-1, Y);

i=i+2*l;

printf("/nMPASS(): i=%d", i);

}

printf("/nMPASS(): i+l-1=%d, n=%d", i+l-1, n);

// merge remaining file of length < 2l

if(i+l-1<n){// "i+l-1"正好是下一个分区的中间值m

// 如果m(=i+l-1)<n, 说明剩下的未归并分区中的数目多于“归并长度的一半”（即l/2）

// 则还需要进行一次归并

MERGE(X, i, i+l-1, n, Y);

}else{

// 否则，不需要再排序，因为这里只有不到本次归并长度一般的数目，显然已经在上一次归并循环中排好序了

// 简单的拷贝即可，等待下一次2倍归并长度的排序

for(int t=i; t<=n; ++t){

Y[t] = X[t];

}

}

Print(Y, n);

}

void MSORT(int* X, int n)

{

int* Y = (int *)malloc(sizeof(int)*(n+1)/*Guard 1*/);

int l = 1;

printf("/nStart sorting!");

Print(X, n);

while(l<n){

MPASS(X, Y, n, l);

l = 2*l;

MPASS(Y, X, n, l);

l = 2*l;

}

free(Y);

printf("/nFinished!");

}

int main(){

int A[] = {0, 435, 6};

int B[] = {0, 7, 7435, 23, 66, 32, 75, 45, 886, 3456, 34234, 32, 45, 78, 8, 78, 34, 79, 567, 17, 34, 754, 345, 78, 9};

MSORT(A, sizeof(A)/sizeof(A[0])-1);

//MSORT(B, sizeof(B)/sizeof(B[0])-1);

return 0;

}

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