Trapping Rain Water

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;
	}

Trapping Rain Water,布布扣,bubuko.com

Trapping Rain Water

原文：http://blog.csdn.net/cow__sky/article/details/21038031