原题地址:https://oj.leetcode.com/problems/trapping-rain-water/ 题意: 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. For example, Given [0,1,0,2,1,0,1,3,…

Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining. Note: Both m and n are less than 110. The height of each unit cell is greater th…

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. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map is represented by a…

https://oj.leetcode.com/problems/trapping-rain-water/ Trapping Rain WaterGiven 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. For example, Given [0,1,0,…

Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining. Note: Both m and n are less than 110. The height of each unit cell is greater th…

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. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map is represented by a…

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. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map is represented by a…

题意 题目 思路 我一开始想的时候只考虑到一个结点周围的边界的情况,并没有考虑到边界的高度其实影响到所有的结点盛水的高度. 我们可以发现,中间是否能够盛水取决于边界是否足够高于里面的高度,所以这必然是一个从外到内,从小到大的一个过程.因为优先队列必然首先访问的是边界中最小的高度,如果周围小于这个高度那么必然是存在可以盛水的地方,就算是没有也没有任何的损失,只是更新了高度,但是队列依然会从高度小的结点开始访问: 实现 typedef struct node { int x, y; node(int…

class Solution(object): def trap(self,nums): leftmosthigh = [0 for i in range(len(nums))] leftmax=0 for i in range(len(nums)): if nums[i] > leftmax: leftmax=nums[i] leftmosthigh[i] = leftmax print leftmosthigh sums=0 rightmax=0 for i in reversed(rang…

LeetCode 42. Trapping Rain Water Python解法 解题思路: 本思路需找到最高点左右遍历,时间复杂度O(nlogn),以下为向左遍历的过程. 将每一个点的高度和索引存成一个元组 (val, idx) 找到最高的点(可能有多个,任取一个),记为 (now_val, now_idx). 向左找第一个val不大于now_val的点(left_val, left_idx). 以left_val作为水平面,用height[left_val+1, now_val-1]中每一…