这题放上来是因为自己第一回见到这种题,觉得它好玩儿 =)
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!
class Solution { public: int trap(int A[], int n) { } };
题目不难,观察下就可以发现被水填满后的形状是先升后降的塔形,因此,先遍历一遍找到塔顶,然后分别从两边开始,往塔顶所在位置遍历,水位只会增高不会减小,且一直和最近遇到的最大高度持平,这样知道了实时水位,就可以边遍历边计算面积。
从看题目开始一共十分钟,算本菜鸟在AC率二字打头的题目中至今最快的一次。。
class Solution { public: int trap(int A[], int n) { if(n <= 2) return 0; int max = -1, maxInd = 0; int i = 0; for(; i < n; ++i){ if(A[i] > max){ max = A[i]; maxInd = i; } } int area = 0, root = A[0]; for(i = 0; i < maxInd; ++i){ if(root < A[i]) root = A[i]; else area += (root - A[i]); } for(i = n-1, root = A[n-1]; i > maxInd; --i){ if(root < A[i]) root = A[i]; else area += (root - A[i]); } return area; } };
总结:
这道题和LeetCode上 "Candy" 一题都采用了定义两个指针向中部某一点靠拢的做法(当然Candy还有更快的解,见以前的这篇),这也算是小技巧之一吧,在需要时要能第一时间想到。