这是一个很有意思的问题,求解最大容积问题,值得动脑筋想一想。
原题如下:
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!
图中浅颜色的就是雨水的含量,从图中我们不难发现,在每一列有水的雨水含量就是这一列左右两侧最高的两个柱子中的较矮的柱子和当前列的高度差。这样我们就能逐次计算出来分列的含量。当然,我们可以在计算每一列的时候都计算一遍两侧的最大值,但是这样是浪费时间的,我们可以分治解决,先求解出最高的柱子,这样这一柱子在一定时刻内可以当做其中一侧的最值,逐渐缩小范围即可。下面是自己写的代码,可能比较乱,但是思想还是可以感受一下的,在最后会贴出来大牛们写的高质量代码。<span style="font-size:18px;">class Solution { public: int max_value(int a,int b) { return a>b?a:b; } int max_element_position(int A[],int begin,int end) { int max = A[begin]; int pos = begin; for(int i = begin+1;i<=end;i++) if(A[i]>max) { max = A[i]; pos = i; } return pos; } int water(int A[],int start,int left_max,int end,int right_max) { int left = 0,right = 0,mid = 0; if(start == end) { if(left_max>A[start] && right_max>A[start]) return min(left_max,right_max) - A[start]; else return 0; } int max = max_element_position(A,start,end); if(A[max]<left_max && A[max]<right_max)//这里要注意,中间的最大值也可能上面存储水 { mid = min(left_max,right_max) - A[max]; } if(max-1 >= start) left = water(A,start,left_max,max-1,max_value(right_max,A[max])); if(max+1 <= end) right = water(A,max+1,max_value(left_max,A[max]),end,right_max); return left+right+mid; } int trap(int A[], int n) { if(n<=2) return 0; return water(A,0,0,n-1,0); } };</span>下面的方法确实是很理解了很多,不错不错。
int trap(int A[], int n) { // Note: The Solution object is instantiated only once. if(A==NULL || n<1)return 0; int maxheight = 0; vector<int> leftMostHeight(n); for(int i =0; i<n;i++) { leftMostHeight[i]=maxheight; maxheight = maxheight > A[i] ? maxheight : A[i]; } maxheight = 0; vector<int> rightMostHeight(n); for(int i =n-1;i>=0;i--) { rightMostHeight[i] = maxheight; maxheight = maxheight > A[i] ? maxheight : A[i]; } int water = 0; for(int i =0; i < n; i++) { int high = min(leftMostHeight[i],rightMostHeight[i])-A[i]; if(high>0) water += high; } return water; }
每日算法之三十三:Trapping Rain Water,布布扣,bubuko.com
原文:http://blog.csdn.net/yapian8/article/details/30544939