题目要求: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!
代码如下:
class Solution { public: int trap(int A[], int n) { int maxIdx = 0; int water = 0; //找到最长的木板,设为maxIdx for(int i = 1; i < n; i++){ if(A[i] > A[maxIdx]){ maxIdx = i; } } int max = A[0]; //左侧逼近 for(int i = 1; i < maxIdx; i++){ if(max < A[i]) max = A[i]; //木板左边(max)和右边(最高)都比它高,则可以放 // max - 该木板长度 的水 else water += max - A[i]; } //右侧逼近 max = A[n - 1]; for(int i = n - 2; i >= maxIdx; i--){ if(max < A[i]) max = A[i]; else water += max - A[i]; } return water; } };
时间: 2024-11-12 16:04:42