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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; }
42. Trapping Rain Water
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