Givennnon-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], return6. 题目大概意思为有
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 Queue 1 2 3 4 5 6 7 8 9 10...
class Solution: def trapRainWater(self, heightMap: List[List[int]]) -> int: import heapq if not heightMap: return 0 heap = [] cur_max = float("-inf") visited = set() row = len(heightMap) col = len(heightMap[0]) res = 0 # 边界放进去 #行 for j in range(col): heapq.hea...
struct Node{ int i,j,h; Node(int ii,int jj,int hh):i(ii),j(jj),h(hh){} bool operator <(const Node &root) const{ return h>root.h; } }; class Solution { public: int trapRainWater(vector<vector<int>>& heightMap) { if(heightMap.size()==0) return 0; int m=heightMap....
int trapRainWater(vector<vector<int>>& heightMap) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2 heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] 1 2 [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] [[3,3,3,3,3],[3,2,...
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0407 Trapping Rain Water II 47.4% Hard 0408 Valid Word Abbreviation 34.8% Easy 0409 Longest Palindrome Go 54.6% Easy 0410 Split Array Largest Sum Go 53.2% Hard 0411 Minimum Unique Word Abbreviation 39.2% Hard 0412 Fizz Buzz Go 68.6% Easy 0413 Arithmetic Slices Go 65.0% Medium 0414...
Leetcode 407. Trapping Rain Water II Problem: Given anm x nmatrix 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: Bothmandnare less than 110. The height of each unit cell is ...
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