TrappingRainWater

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。

 

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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;     }   }