[Leetcode] 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!

Pasted from <http://leetcode.com/onlinejudge>

Analysis: 


Find the max height at i, then the trapped water at position j less than (at left of) i depends on its max height on the left of i; similarly, the trapped water at position j greater than (at the right of) i depends on its max height on the right of j. No extra space needed for the solution.

int trap(int A[], int n) {
    int maxHeight = 0;
    int maxHeightPos = 0;
    for (int i=0; i maxHeight) {
            maxHeight = A[i];
            maxHeightPos = i;
        }
    }

    int sum = 0;
    int leftHeight = 0;
    for (int i=0; i < maxHeightPos; i++)
    {
        sum += max (0, leftHeight - A[i]);
        leftHeight = max (A[i], leftHeight);
    }

    int rightHeight = 0;
    for (int i=n-1; i > maxHeightPos; i--)
    {
        sum += max (0, rightHeight - A[i]);
        rightHeight = max (A[i], rightHeight);
    }

    return sum;
}