19.2.4 [LeetCode 42] Trapping Rain Water
阿新 • • 發佈:2019-02-04
continue pin water 沒有 tor open bar solution array
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
題解
我覺得這題還挺難的,是我還沒有深刻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 ifView Code(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 };
19.2.4 [LeetCode 42] Trapping Rain Water