合并两个已排序的数组为同一个数组

问题描述: 设子数组a[0:k]和a[k+1:n-1]已排好序（0<=k<=n-1).试设计一个合并这两个子数组为排好序的数组a[0:n-1]的算法．要求算法在最坏的情况下所用的计算时间为O(n), 且只用到O(1)的辅助空间． #include <stdio.h>

void DisplayArray(int *pArray, int nLen)

{

    for (int i = 0; i < nLen; ++i)

    {

        printf("array[%d] = %d\n", i, pArray[i]);

    }

}

// pArray1和pArray2是已经排好序的数组,要求将它们按照顺序合并到pArray中

// 排序之后的数组不会有重复的元素

void MergeArray(int *pArray1, int nLen1, int *pArray2, int nLen2, int *pArray)

{

    int i, j, n;

    i = j = n = 0;

    while (i < nLen1 && j < nLen2)                  // 循环一直进行到拷贝完某一个数组的元素为止

    {

        if (pArray1[i] < pArray2[j])                // 拷贝array1的元素

        {

            pArray[n++] = pArray1[i++];

        }

        else if (pArray1[i] > pArray2[j])            // 拷贝array2的元素

        {

            pArray[n++] = pArray2[j++];                      

        }

        else                                          // 相等的元素拷贝

        {

            pArray[n++] = pArray2[j++];                      

            ++i;

        }

    }

    if (i == nLen1)                              // 如果array1已经被拷贝完毕就拷贝array2的元素

    {

        while (j < nLen2)

            pArray[n++] = pArray2[j++];

    }

    else                                         // 如果array2已经被拷贝完毕就拷贝array1的元素

    {

        while (i < nLen1)

            pArray[n++] = pArray1[i++];

    }

}

int main()

{      

    int array1[] = {1, 4, 5, 7};

    int array2[] = {2, 3, 6, 8};

    int array3[8];

    MergeArray(array1, 4, array2, 4, array3);

    printf("Merge Array:\n");

    DisplayArray(array3, 8);

    return 1;

}