trapping rain water

题目;

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!

解答：

找到最长的那块木板，假设其下标为maxIdx。

分别从左侧和右侧向其逼近。

在左侧逼近过程中：

如果一个木板的长度小于已经遍历的最大长度max，即max>该木板<maxIdx，所以在该木板位置能存max - 该木板长度的水量（左右两侧各有一个木板长于它）。

如果一个木板的长度大于已经遍历的最大长度max，即max<该木板<maxIdx，所以在该木板位置不能存水（因为左右两侧只有一个木板（maxIdx）长于它）。更新max值。

右侧逼近过程与左侧相似。

代码：

class Solution {public:    int trap(int A[], int n) {        int i;int maxline, max;int water = 0;maxline = 0;max = A[0];for(i = 1; i < n; i++){if(A[i] > max){max = A[i];maxline = i;}}max = A[0];for(i = 1; i < maxline; i++){if(A[i] < max){water += max - A[i];}else{max = A[i];}}max = A[n-1];for(i = n-2; i > maxline; i--){if(A[i] < max){water += max - A[i];}else{max = A[i];}}return water;    }};