42. Trapping Rain Water

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!

 

 

此题与给小孩分糖那题很相似，都是可以左边遍历一遍，右边遍历一遍。

public int Trap(int[] height) {
        int max =0;
        int res = 0;
        int size = height.Count();
        if(size <= 2) return res;
        var maxArr =new int[size];
        for(int i = 0;i< height.Count();i++)
        {
            maxArr[i] =Math.Max(height[i],max);
            max = Math.Max(max,height[i]);
        }
        int rightMost  =0;
        int lessHeight = 0;
        for(int i = size -1 ;i>=0;i--)
        {
            rightMost = Math.Max(rightMost, height[i]);
            lessHeight = Math.Min(rightMost, maxArr[i]);
            res +=  (lessHeight - height[i] > 0)?lessHeight - height[i] : 0;
        }
        return res;
    }

 

参考OJ上大神的做法，用stack只存比peek值小的元素的index，因为index要用来算两个边之间的宽度。这个方法是将水分层了，也就是比如上图test case中间存水的部分，因为先出现1的右边边界，先加上1的水，后来又遇到3的边界，左边为1时没有水加入，左边为2时加入水，水的高度为更低的边与基地的差（Min（2,3）-1）。

 

 public int Trap(int[] height) {
        int res = 0;
        int size = height.Count();
        if(size <= 2) return res;
        int i =0;
        var stack = new Stack<int>();//store the largerst heigth index
        while(i< size)
        {
            if(stack.Count()==0 || height[i] <= height[stack.Peek()]) stack.Push(i++);
            else
            {
                int bot = stack.Pop();
                res += (stack.Count()==0)?0:((Math.Min(height[stack.Peek()], height[i]) - height[bot])*(i - stack.Peek()-1));; 
            }
        }
        return res;
    }