Leetcode 407. Trapping Rain Water II 收集雨水2 解题报告
2016-09-25 21:54
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1 解题思想
我看了下题目,发现比预想中的简单,加之我比较烂,所以其实我还是没做,只是看懂了上回贴的代码,然后做了一下注释,现在我来讲下题目。首先请看下上一题,上一题是2D的这题是3D的:
Leetcode #42. Trapping Rain Water 雨水收集 解题报告
题目的意思,就是给定了一个矩阵,代表一些木头桩,高度不一样,所以下雨的时候能够保存住一些水,现在问说能够收集多少雨水?
所谓的能收集雨水,就是要中间低,四边高,围起来才行,这就是核心idea和题目42一样。
所以这题的思想是:
1、构造一个优先队列,每次都取出高度最矮的一个。
2、首先将四周的桩都加入,因为记得四周是无法盛水的
3、每次从优先队列中取出一个最矮的cell,若他比未访问过的四周的四个高了delta h,那么总的盛水量多加delta h,否则不加,注意四周只要一访问过下次就不能访问了。
因为若当前cell被取出,且其某个邻居cell-n高度低于cell,那么可以得出:cell-n的邻居里面最矮的就是cell,所以cell可以决定cell-n的盛水量。
2 原题
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 than 0 and is less than 20,000.
Example:
Given the following 3x6 height map:
[
[1,4,3,1,3,2],
[3,2,1,3,2,4],
[2,3,3,2,3,1]
]
Return 4.
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 are trapped between the blocks. The total volume of water trapped is 4.
3 参考代码
修改于于:https://discuss.leetcode.com/topic/60371/java-versionpublic class Solution { public int trapRainWater(int[][] heightMap) { //一个单元格用一个Cell来表示 class Cell{ int x, y,h; Cell(int x, int y, int height){ this.x = x; this.y = y; h = height; } } if (heightMap == null || heightMap.length == 0 || heightMap[0].length == 0) { return 0; } int m = heightMap.length; int n = heightMap[0].length; //优先队列,每次按照优先度输出队列,而不是按照顺序,这里是每次输出最矮的哪一个 PriorityQueue<Cell> pq = new PriorityQueue<>((v1,v2)->v1.h - v2.h); boolean[][] visited = new boolean[m] ; //将四周初始化为访问过的,周围的一边是怎么都没法盛水的 for(int i = 0; i < n; i++){ visited[0][i] = true; visited[m-1][i] = true; pq.offer(new Cell(0, i, heightMap[0][i])); pq.offer(new Cell(m-1, i, heightMap[m-1][i])); } for(int i = 1; i < m-1; i++){ visited[i][0] = true; visited[i][n-1] = true; pq.offer(new Cell(i, 0, heightMap[i][0])); pq.offer(new Cell(i, n-1, heightMap[i][n-1])); } //四个方向 int[] xs = {0, 0, 1, -1}; int[] ys = {1, -1, 0, 0}; int sum = 0; //开始计算收集到的雨水,每次取出符合条件最矮的按个,然后计算差值,就是当前单元格可以容纳的了 while (!pq.isEmpty()) { Cell cell = pq.poll(); for (int i = 0; i < 4; i++) { int nx = cell.x + xs[i]; int ny = cell.y + ys[i]; if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny]) { visited[nx][ny] = true; sum += Math.max(0, cell.h - heightMap[nx][ny]); pq.offer(new Cell(nx, ny, Math.max(heightMap[nx][ny], cell.h))); } } } return sum; } }
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