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19.2.4 [LeetCode 42] Trapping Rain Water

时间:2019-02-04 15:32:07      阅读:203      评论:0      收藏:0      [点我收藏+]

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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.

技术图片
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

技术图片
 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 };
View Code

 

19.2.4 [LeetCode 42] Trapping Rain Water

标签:load   example   cos   rap   onclick   set   view   upload   src   

原文地址:https://www.cnblogs.com/yalphait/p/10351713.html

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