合并数组的两个已排序的一半

时间:2017-09-21 01:00:16

标签: c++ arrays sorting mergesort

我试图在C ++中实现这个mergeSort函数,但我收效甚微。到目前为止我创建的算法对数组的两半进行了排序,但是,我知道在返回之前我需要一些代码,在后半部分之后需要将数组的两半合并在一起。这是我遇到一些问题的地方,我无法弄清楚如何将两者合并在一起。

int mergeSort(empType A[], int first, int last)
{

int mid = (first + last + 1) / 2;       // get midpoint
int comp = 0;                       // holds # of comparisons
empType tempArray;

// Loop to middle, start at first + 1
for (int i = first + 1; i < mid; i++)
{
    // if given score < previous
    if (A[i].Score < A[i - 1].Score)
    {
        // set j = what i is at in original for loop
        // loop to first + 1 and decrement
        for (int j = i; j > first + 1; j--)
        {
            // if element's score < previous
            if (A[j].Score < A[j - 1].Score)
            {
                tempArray.ID = A[j].ID;             // Put ID into tempArray
                tempArray.Score = A[j].Score;       // Put Score into tempArray
                A[j].ID = A[j - 1].ID;              // Move on to next ID element
                A[j].Score = A[j - 1].Score;        // Move on to next Score element
                A[j - 1].ID = tempArray.ID;         // Put next ID into tempArray
                A[j - 1].Score = tempArray.Score;   // Put next Score into tempArray
                comp++;                             // Increase computations
            }
        }
    }
}
// loop from middle to last elements
for (int i = mid; i < last; i++)
{
    // if current score < previous
    if (A[i].Score < A[i - 1].Score)
    {
        for (int j = i; j > mid + 1; j--)
        {
            if (A[j].Score < A[j - 1].Score)
            {
                tempArray.ID = A[j].ID;
                tempArray.Score = A[j].Score;
                A[j].ID = A[j - 1].ID;
                A[j].Score = A[j - 1].Score;
                A[j - 1].ID = tempArray.ID;
                A[j - 1].Score = tempArray.Score;
                comp++;
            }
        }
    }
}




return comp;    // return the number of comparisons
}

0 个答案:

没有答案
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