Trapping Rain Water

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.

public class Solution {  
    public int trap(int[] height) {
        if (height == null || height.length == 0) {
            return 0;
        }

        int[] left = new int[height.length];
        left[0] = height[0];
        int max = height[0];

           for (int i = 1; i < height.length; i++) {
               max = Math.max(max, height[i]);
               left[i] = max;
           }

           int[] right = new int[height.length];
           right[height.length - 1] = height[height.length - 1];
           max = height[height.length - 1];

           for (int i = height.length - 2; i >= 0; i--) {
               max = Math.max(max, height[i]);
               right[i] = max;
           }

           int sum = 0;
           for (int i = 0; i < height.length; i++) {
               int min = Math.min(left[i], right[i]);

               if (min - height[i] > 0) {
                   sum += min - height[i];
               }
           }

           return sum;
    }
}

Hope this helps,
Michael