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问题描述：
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.
算法分析：观察下就可以发现被水填满后的形状是先升后降的塔形，因此，先遍历一遍找到塔顶，然后分别从两边开始，往塔顶所在位置遍历，水位只会增高不会减小，且一直和最近遇到的最大高度持平，这样知道了实时水位，就可以边遍历边计算面积。
public class TrapingRainWater
{
public int trap(int[] height)
{
int n = height.length;
if(n <= 2)
{
return 0;
}
int max = -1;
int maxIndex = 0;
for(int i = 0; i < n; i ++)
{
if(height[i] > max)
{
max = height[i];
maxIndex = i;
}
}
int area = 0;
int root = height[0];
for(int i = 0; i < maxIndex; i ++)
{
if(root < height[i])
{
root = height[i];
}
else
{
area += (root - height[i]);
}
}
root = height[n-1];
for(int i = n-1; i > maxIndex; i --)
{
if(root < height[i])
{
root = height[i];
}
else
{
area += (root - height[i]);
}
}
return area;
}
}
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原文地址：http://www.cnblogs.com/masterlibin/p/5741920.html
