問題描述

將兩個有序的順序表合併成一個新的有序順序表，由函式返回結果順序表

演算法思想

本題實際就是歸併排序的一種特殊情況，因為兩個順序表皆有序，這樣我們只需要不斷的取下兩個順序表中表頭元素較小的那個數，然後將其存入新的順序表，最後看哪個表還有剩餘，將剩下的部分直接新增到新的順序表後面即可。

演算法描述

SqList* Merge(SqList *L1, SqList *L2)
{
    SqList *L = (SqList*)malloc(sizeof(SqList));
    int i = 0, j = 0, k = 0;
    while(i<L1->length&&j<L2->length){  
        if
 
 (L1->data[i] <= L2->data[j]){
            L->data[k++] = L1->data[i++];
        }
        else{
            L->data[k++] = L2->data[j++];
        }
    }
    while (i != L1->length){
        L->data[k++] = L1->data[i++];
    }
    while (j != L2->length){
        L->data[k++] = L1->data[j++];
    }
    L->length = k;
    return
 
 L;
}

具體程式碼見附件

附件

#include<stdio.h>
#include<stdlib.h>
#define MaxSize 100
typedef int ElemType;

typedef struct{
    ElemType data[MaxSize];
    int length;
}SqList;

SqList* Merge(SqList *, SqList *);
void print(SqList*);

int main(int argc, char* argv[])
{
    SqList SL1, SL2;
    SL1.length = 10
 
;
    SL2.length = 10;
    for (int i = 0; i < SL1.length; i++){
        SL1.data[i] = 2 * i + 1;
    }
    for (int i = 0; i < SL2.length; i++){
        SL2.data[i] = 2 * i;
     }
    print(&SL1);
    print(&SL2);
    SqList *SL;
    SL = Merge(&SL1, &SL2);
    print(SL);
    return 0;
}

SqList* Merge(SqList *L1, SqList *L2)
{
    SqList *L = (SqList*)malloc(sizeof(SqList));
    int i = 0, j = 0, k = 0;
    while(i<L1->length&&j<L2->length){  
        if (L1->data[i] <= L2->data[j]){
            L->data[k++] = L1->data[i++];
        }
        else{
            L->data[k++] = L2->data[j++];
        }
    }
    while (i != L1->length){
        L->data[k++] = L1->data[i++];
    }
    while (j != L2->length){
        L->data[k++] = L1->data[j++];
    }
    L->length = k;
    return L;
}

void print(SqList *L){
    for (int i = 0; i < L->length; i++){
        printf("%4d", L->data[i]);
    }
    printf("\n");
}