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.

参考：

http://blog.csdn.net/doc_sgl/article/details/12307171

public class Solution {
    public int trap(int[] A) {
        int waterSum = 0;
        int leftMaxHeight[] = new int[A.length];
        int rightMaxHeight[] = new int[A.length];
        
        int maxHeight = 0;
        for(int i = 0; i < A.length; i++){
            leftMaxHeight[i] = maxHeight;
            maxHeight = maxHeight > A[i] ? maxHeight : A[i];
        }//for
        
        maxHeight = 0;
        for(int i = A.length - 1; i >= 0; i--){
            rightMaxHeight[i] = maxHeight;
            maxHeight = maxHeight > A[i] ? maxHeight : A[i];
        }//for
        
        for(int i = 0; i < A.length; i++){
            int minHeight = Math.min(leftMaxHeight[i], rightMaxHeight[i]);
            int elementSum = minHeight - A[i];
            if(elementSum > 0)
                waterSum += elementSum;
        }//for
        
        return waterSum;
    }
    
}