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19.2.4 [LeetCode 42] Trapping Rain Water

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.

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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

技術分享圖片
 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 };
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19.2.4 [LeetCode 42] Trapping Rain Water