19.2.4 [LeetCode 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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

题解

我觉得这题还挺难的，是我还没有深刻get的思想

 1 class Solution {
 2 public:
 3     int trap(vector<int>& height) {
 4         int ans = 0, size = height.size(), i = 0;
 5         stack<int>q;
 6         while (i < size) {
 7             if (q.empty()||height[i]<=height[q.top()]) {
 8                 q.push(i++);
 9                 continue;
10             }
11             else {
12                 int now = q.top(); q.pop();
13                 if (q.empty())continue;
14                 ans += (min(height[i], height[q.top()]) - height[now])*(i - q.top() - 1);
15             }
16         }
17         return ans;
18     }
19 };

标签：map,elevation,Trapping,19.2,int,42,height,water,size
来源： https://www.cnblogs.com/yalphait/p/10351713.html