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[n] 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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