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!

思路：

找到最高的一个板子，然后从两边往中间靠近，如果高度小于当前的最大高度，结果加上当前最大高度-板子高度，否则不存水，只更新最大高度即可。

代码：

    int trap(int A[], int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int i;
        int curmax, max = INT_MIN, maxIndex, result = 0;
        if(n == 0)
            return 0;
        for(i = 0; i < n; i++){
            if(A[i] > max){
                max = A[i];
                maxIndex = i;
            }
        }
        curmax = A[0];
        for(i = 1; i < maxIndex; i++){
            if(A[i] < curmax){
                result += (curmax-A[i]);
            }
            else{
                curmax = A[i];
            }
        }
        curmax = A[n-1];
        for(i = n-2; i > maxIndex; i--){
            if(A[i] < curmax){
                result += (curmax-A[i]);
            }
            else{
                curmax = A[i];
            }
        }
        return result;
    }