算法讲解: http://www.geeksforgeeks.org/merge-sort/

测试题

Merge Sort (Function Problem)
The task is to complete merge() function which is used to implement Merge Sort.

Input:
First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.

Output:
Sorted array in increasing order is displayed to the user.

Constraints:
1 <=T<= 50
1 <=N<= 1000
1 <=arr[i]<= 1000

Example:

Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1

Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10

C++

/*
Please note that it's Function problem i.e.
you need to write your solution in the form of Function(s) only.
Driver Code to call/invoke your function would be added by GfG's Online Judge.*/


/* The task is to complete merge() which is used
   in below mergeSort() */
/*   l is for left index and r is right index of the
   sub-array of arr to be sorted
void mergeSort(int arr[], int l, int r) {
    if (l < r)   {
        int m = l+(r-l)/2;
        mergeSort(arr, l, m);
        mergeSort(arr, m+1, r);
        merge(arr, l, m, r);
    }
} */
// Merges two subarrays of arr[].  First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
     // Your code here
     
     int arr_len = r-l+1;
     int arr_tmp[arr_len];
     
     int i = 0; //equal l-l
     int j = m+1-l;
     
     for(int s = 0; s<=r-l; s++)
     {
        arr_tmp[s] = arr[s+l];
     }
     
     int k = l;
     while(i<=m-l && j<=r-l)
     {
         if(arr_tmp[i]<=arr_tmp[j])
         {
             arr[k] = arr_tmp[i];
             i++;
         }
         else
         {
             arr[k] = arr_tmp[j];
             j++;
         }
         k++;
     }
     while(i<=m-l)
     {
         arr[k] = arr_tmp[i];
         i++;
         k++;
     }
     while(j<=r-l)
     {
         arr[k] = arr_tmp[j];
         j++;
         k++;
     }
}