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!
解题：
设置两个指针从两边往中间走，每次把min(A[left], A[right])作为下界墙，以left, right为两边计算当前层的总面积，迭代下一层。
最后将算起来的总面积减去黑块的面积。
class Solution {
public:
int trap(int A[], int n) {
if(!n) return 0;
area = tot = 0;
int le = 0, ri = n - 1;
for(;le < n && !A[le]; le ++);
for(;ri >= le && !A[ri]; ri --);
for(int i = le; i <= ri; i ++)
tot += A[i];
gao(A, 0, n - 1, 0);
return area - tot;
}
void gao(int A[], int le, int ri, int pre) {
while(A[le] <= pre && le < ri) le ++;
while(A[ri] <= pre && le < ri) ri --;
if(le == ri) {
tot -= A[le] - pre;
return ;
}
area += (min(A[le], A[ri]) - pre) * (ri - le + 1);
gao(A, le, ri, min(A[le], A[ri]));
return ;
}
private:
int area, tot;
};