After the rain, water is trapped between the blocks. The total volume of water trapped is 4. 42. Trapping Rain Water的拓展,由2D变3D了。解法跟之前的完全不同了,之前那道题由于是二维的,我们可以用双指针来做,而这道三维的,我们需要用BFS来做。 Java: Priority
The above image represents the elevation map[[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]before the rain. After the rain, water is trapped between the blocks. The total volume of water trapped is 4. 解题思路: 类比于trapped rain water, 上一个是一维的数组,而这一个是二维的...
After the rain, water are trapped between the blocks. The total volume of water trapped is 4. 这道题是之前那道Trapping Rain Water的拓展,由2D变3D了,感觉很叼。但其实解法跟之前的完全不同了,之前那道题由于是二维的,我们可以用双指针来做,而这道三维的,我们需要用BFS来做,解法思路很巧妙,下面我们...
## LeetCode 42H - Trapping Rain Water,双指针法 from typing import List class Solution: ## height 表示列表中元素的取值,高度 def trap(self, height: List[int]) -> int: if not height: ## 如果没有任何高度 return 0 ## 接不住任何雨水 ## 初始化左右指针 l, r = 0, len(height) - 1...
https://leetcode-cn.com/problems/trapping-rain-water/ 题目内容 给定n 个非负整数表示每个宽度为 1 的柱子的高度图,计算按此排列的柱子,下雨之后能接多少雨水。 示例图 示例: 输入: [0,1,0,2,1,0,1,3,2,1,2,1] 输出: 6 思路 我们可以设置两个数组,「left_height_array」和「right_height_array...
https://leetcode-cn.com/problems/trapping-rain-water/ 题目内容 给定n 个非负整数表示每个宽度为 1 的柱子的高度图,计算按此排列的柱子,下雨之后能接多少雨水。 示例: 输入: [0,1,0,2,1,0,1,3,2,1,2,1] 输出: 6 思路 我们可以设置两个数组,「left_height_array」和「right_height_array」; ...
思路2 接雨水嘛,两边比中间高就能接到,那么我只需要求出所有比左右两边高的点,他们两两组合就成一个可以接到水的坑,以原题中的示例1为例,下标为1,3,7,10的点就是我们所求。代码如下 /** * @param {number[]} height * @return {number} ...
leetcode 42 接雨水 trapping-rain-water【ct】,思路:当前点的=min(lmax,rmax)-cur,最终加到一起
[LeetCode] 42. Trapping Rain Water 接雨水 2023-11-18 02:20:3131:53 39 所属专辑:LeetCode算法题目讲解 喜欢下载分享 声音简介[LeetCode] 42. Trapping Rain Water 接雨水博客园:https://www.cnblogs.com/grandyang/p/4402392.htmlGitHub:https://github.com/grandyang/leetcode/issues/42个人网页:https:...
LeetCode42 Trapping Rain Water this problem is really classic and we have a really classic way to solve this problem: we iterate every value of given array, and each time, for a fixed value, we only consider the water it can contained, and we maintained two variables: leftMax and right...