leetcode_Trapping Rain Water

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.

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!

思路：

动态规划题目

1.从左至右算出每个点的左方的最大值

2.从右至左算出每个点的右方的最大值

3.从左至右循环sum+=min(leftMax[i]+rightMax[i])-arr[i]

4.sum的值就是所能储存的最大水量

代码：

public int trap(List<Integer> height) {        int count=0;        if(height==null)            return 0;        if(height.size()==0)            return 0;        int len=height.size();        int arrLeft[]=new int[len];        int arrRight[]=new int[len];        int max=0;        int heigh=0;        for(int i=0;i<len;i++)        {            heigh=height.get(i);            if(heigh>max)                max=heigh;            arrLeft[i]=max;        }        max=0;        heigh=0;        for(int i=len-1;i>=0;i--)        {            heigh=height.get(i);            if(heigh>max)                max=heigh;            arrRight[i]=max;        }        int newLen=len-1;        int min=0;        for(int i=1;i<newLen;i++)        {            heigh=height.get(i);            min=arrLeft[i]<arrRight[i]?arrLeft[i]:arrRight[i];            if(min>heigh)                count+=min-heigh;        }        return count;    }