合并两个数组

#include <iostream>

using namespace std;

int sortedInsert(int A[],int lenA,int B[],int lenB)

{

    if(B == NULL)

        return lenA;

    int len = lenA + lenB;

    int indexNew = len-1;

    int indexA = lenA-1;

    int indexB = lenB-1;

    while( indexA>=0 && indexB>=0 )

    {

        if(A[indexA] > B[indexB])

        {

            A[indexNew] = A[indexA--];

            

        }else{

            A[indexNew] = B[indexB--];    

        }

    //    printf("%d \n",A[indexNew]);

        //printf("here lenA = %d,lenB = %d\n",indexA,indexB) ;

        indexNew --;

    }

    //下面处理剩余的那部分;

    //printf("here lenA = %d,lenB = %d\n",indexA,indexB) ;

    while(indexA>=0)

    {

        //

        A[indexNew] = A[indexA];

        //printf("%d ",A[indexNew]);

        indexNew--; indexA -- ;

    } 

    //printf("OK");

    while(indexB>=0)

    {

        

        A[indexNew] = B[indexB];

        //printf("%d ",A[indexNew]);

        indexNew--; indexB -- ;

    } 

    return lenA+lenB;

}

int intCmp(int * A ,int lenA,int * expected ,int lenExpected)

{

    if(lenA != lenExpected)

        return -1;

    for(int i = 0;i<lenA;i++)

        if(A[i]!=expected[i])

        {

            //printf("%d %d %d\n",i,A[i],expected[i]);

            return -1;

        }

    return 0;

}

void test(char * testNumber,int *A,int lenA,int *B,int lenB,int* expected,int lenExpected)

{

    lenA = sortedInsert(A,lenA,B,lenB);

    printf("%s ",testNumber); 

    if(intCmp(A,lenA,expected,lenExpected) == 0 )

        printf("passed\n");

    else

        printf("failed\n");

}

void test1()//B插在A的后头; 

{

    int A[100]={1,3,5,7,9};

    int B[]={2,4,6,8};

    //sortedInsert(A,5,B,4);

    int expected[]={1,2,3,4,5,6,7,8,9};

    test("test1",A,5,B,sizeof(B)/sizeof(int),expected,sizeof(expected)/sizeof(int));

} 

void test2()//B要插在A的前头; 

{

    int A[100]={6,7,8,9};

    int B[]={1,2,3,4,5};

    //sortedInsert(A,5,B,4);

    int expected[]={1,2,3,4,5,6,7,8,9};

    test("test2",A,4,B,sizeof(B)/sizeof(int),expected,sizeof(expected)/sizeof(int));

} 

void test3()//A没有内容 

{

    int A[100]={};

    int B[]={1,3,5,7,9};

    //sortedInsert(A,5,B,4);

    int expected[]={1,3,5,7,9};

    test("test3",A,0,B,sizeof(B)/sizeof(int),expected,sizeof(expected)/sizeof(int));

} 

void test4()//B没有内容 

{

    int A[100]={1,3,5,7,9};

    int B[]={};

    //sortedInsert(A,5,B,4);

    int expected[]={1,3,5,7,9};

    test("test4",A,5,B,sizeof(B)/sizeof(int),expected,sizeof(expected)/sizeof(int));

}  

void test5()//A,B都没有内容 

{

    int A[100]={};

    int B[]={};

    //sortedInsert(A,5,B,4);

    int expected[]={};

    test("test5",A,0,B,sizeof(B)/sizeof(int),expected,sizeof(expected)/sizeof(int));

} 

void test6()//B为空指针 

{

    int A[100]={1,3,5,7,9};

    int * B = NULL; 

    //sortedInsert(A,5,B,4);

    int expected[]={1,3,5,7,9};

    test("test6",A,5,B,sizeof(B)/sizeof(int),expected,sizeof(expected)/sizeof(int));

} 

int main()

{

    test1();

    test2();

    test3();

    test4();

    test5();

    test6();

    return 0;

}

 

 合并两个数组（包括字符串）时，如果从前往后复制每个数字（或字符）需要重复移动数字(或字符)多次，那么我们可以考虑从后往前复制，这样就能减少移动的次数，从而提高效率。