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!

class Solution {
public:
/**
* @param heights: a vector of integers
* @return: a integer
*/
int trapRainWater(vector<int> &heights) {
// write your code here
int n = heights.size();
if (n < 1)
{
return 0;
}
int result = 0;
int rightMax
;
rightMax[n-1] = heights[n-1];
for (int i = n-2; i >= 0; i--)
{
if (heights[i] > rightMax[i+1])
{
rightMax[i] = heights[i];
}
else
{
rightMax[i] = rightMax[i+1];
}
}

int begin = 0;
while (begin < n && heights[begin] == 0)
{
begin++;
}

for (int i = begin+1; i < n; i++)
{
if (heights[i] >= heights[begin] || heights[i] == rightMax[i])
{
int top = min(heights[i], heights[begin]);
for (int j = begin+1; j < i; j++)
{
if (heights[j] < top)
{
result += (top - heights[j]);
}
}

begin = i;
}
}

return result;
}
};