leetcode 42. Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

int trap(int* height, int heightSize) {
    if(height==NULL)
        return 0;
	if (heightSize < 3)
		return 0;
	int k = 1, watersum = 0;
	while (k < heightSize-1)
	{
		if (height[k] < height[k - 1])
		{
			int kk = k + 1;
			int endhigh = height[k], endptr=-1;
			while (kk < heightSize&&height[kk] < height[k - 1])
			{
				if (height[kk] > endhigh)
				{
					endptr = kk;
					endhigh = height[kk];
				}
				kk++;
			}
			if (kk == heightSize)
			{
				if (endptr > 0)
				{
					for (int i = k; i < endptr; i++)
						watersum += height[endptr] - height[i];
					k = endptr+1;
				}
				else
					k++;
			}
			else
			{
				for (int i = k; i < kk; i++)
					watersum += height[k - 1] - height[i];
				k = kk+1;
			}
		}
		else
			k++;
	}
	return watersum;
}

accepted