题目原型：
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!
基本思路：
找到最大的数的索引，然后依此为界，分别向左和向右扫描，遇到当前高度比前面leftMaxIndex高度小的则计算面积，反之，更新leftMaxIndex；从右向左，同理！
public int trap(int[] A)
{
if(A.length<=2)
return 0;
int ret = 0;//结果
//找到最大的数的index
int maxIndex = 0;
for(int i = 1;i<A.length;i++)
{
if(A[i]>A[maxIndex])
maxIndex = i;
}
//从左到右遍历计数，知道遇到maxIndex
int leftMax = 0;
for(int i = 1;i<maxIndex;i++)
{
if(A[i]<A[leftMax])
{
ret += (A[leftMax]-A[i]);
}
else
{
leftMax = i;
}
}
//从右到左遍历计数，知道遇到maxIndex
int rightMax = A.length-1;
for(int i = A.length-2;i>maxIndex;i--)
{
if(A[i]<A[rightMax])
{
ret += (A[rightMax]-A[i]);
}
else
{
rightMax = i;
}
}
return ret;
}