Kth Largest Element in an Array

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.

• 很简单的一个问题，把nums排个序，然后输出第k个大的数
• 属于分治部分，就顺便使用并归排序实现

void merge(int *nums, int start, int mid, int end)
{
    int leftSize = mid - start + 1;
    int rightSize = end - mid;
    int i = 0, j = 0;
    int *left = malloc(sizeof(int)* (leftSize + 1));
    int *right = malloc(sizeof(int)* (rightSize + 1) );
    for (int i = 0; i < leftSize; i++)
        left[i] = nums[start + i];    //这里为start+1
    left[leftSize] = -9999;    //设立尾后的标志，这样就可以直接比较大小，而不需要判断是否为空了
    for (int i = 0; i < rightSize; i++)
        right[i] = nums[mid + i + 1];    //注意这里为mid+i+1；mid属于前半部分
    right[rightSize] = -9999;
    for (int k = start; k <= end; k++)
    {
        if (left[i] > right[j])
        {
            nums[k] = left[i];
            i++;
        }
        else
        {
            nums[k] = right[j];
            j++;
        }
    }
    free(left);
    free(right);
}
void mergeSort(int *nums, int start, int end)
{
    int mid = (start + end) / 2;
    if (start < end)
    {
        mergeSort(nums, start, mid);
        mergeSort(nums, mid + 1, end);
        merge(nums, start, mid, end);
    }
}
int findKthLargest(int* nums, int numsSize, int k) {
    mergeSort(nums, 0, numsSize - 1);    //传入的大小按照数组的下标传入
    return nums[k - 1];
}