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