★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:http://www.javashuo.com/article/p-nbaubuxi-me.html
➤若是连接不是山青咏芝的博客园地址,则多是爬取做者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持做者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★html
You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map:git
0
represents the obstacle
can't be reached.1
represents the ground
can be walked through.The place with number bigger than 1
represents a tree
can be walked through, and this positive number represents the tree's height.You are asked to cut off all the trees in this forest in the order of tree's height - always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).github
You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the trees. If you can't cut off all the trees, output -1 in that situation.数组
You are guaranteed that no two trees
have the same height and there is at least one tree needs to be cut off.微信
Example 1:app
Input: [ [1,2,3], [0,0,4], [7,6,5] ] Output: 6
Example 2:this
Input: [ [1,2,3], [0,0,0], [7,6,5] ] Output: -1
Example 3:spa
Input: [ [2,3,4], [0,0,5], [8,7,6] ] Output: 6 Explanation: You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking.
Hint: size of the given matrix will not exceed 50x50.rest
你被请来给一个要举办高尔夫比赛的树林砍树. 树林由一个非负的二维数组表示, 在这个数组中:code
0
表示障碍,没法触碰到.1
表示能够行走的地面.比1大的数
表示一颗容许走过的树的高度.你被要求按照树的高度从低向高砍掉全部的树,每砍过一颗树,树的高度变为1。
你将从(0,0)点开始工做,你应该返回你砍完全部树须要走的最小步数。 若是你没法砍完全部的树,返回 -1 。
能够保证的是,没有两棵树的高度是相同的,而且至少有一颗树须要你砍。
示例 1:
输入: [ [1,2,3], [0,0,4], [7,6,5] ] 输出: 6
示例 2:
输入: [ [1,2,3], [0,0,0], [7,6,5] ] 输出: -1
示例 3:
输入: [ [2,3,4], [0,0,5], [8,7,6] ] 输出: 6 解释: (0,0) 位置的树,你能够直接砍去,不用算步数
提示: 矩阵大小不会超过 50x50 。
Runtime: 7864 ms
1 class Solution { 2 var dirs:[[Int]] = [[0,1],[0, -1],[1, 0],[-1, 0]] 3 func cutOffTree(_ forest: [[Int]]) -> Int { 4 var forest = forest 5 //forest.sort(by:{return $0[2] < $1[2]}) 6 var pq:[[Int]] = [[Int]]() 7 for r in 0..<forest.count 8 { 9 for c in 0..<forest[0].count 10 { 11 var height:Int = forest[r][c] 12 if height > 0 13 { 14 pq.append([r, c, height]) 15 } 16 } 17 } 18 pq.sort(by:{return $0[2] < $1[2]}) 19 var last_i:Int = 0 20 var last_j:Int = 0 21 var result:Int = 0 22 while (!pq.isEmpty) 23 { 24 var new_step:[Int] = pq.removeFirst() 25 var step:Int = bfs(&forest, last_i, last_j, new_step[0], new_step[1]) 26 if step == -1 {return -1} 27 result += step 28 last_i = new_step[0] 29 last_j = new_step[1] 30 } 31 return result 32 } 33 34 func bfs(_ forest:inout [[Int]],_ start_i:Int,_ start_j:Int,_ end_i:Int,_ end_j:Int) -> Int 35 { 36 var queue:[[Int]] = [[Int]]() 37 var m:Int = forest.count 38 var n:Int = forest[0].count 39 var visited:[[Bool]] = [[Bool]](repeating:[Bool](repeating:false,count:n),count:m) 40 queue.append([start_i, start_j]) 41 var level:Int = 0 42 while(!queue.isEmpty) 43 { 44 var size:Int = queue.count 45 for i in 0..<size 46 { 47 var curr:[Int] = queue.removeFirst() 48 if curr[0] == end_i && curr[1] == end_j 49 { 50 return level 51 } 52 for j in 0..<4 53 { 54 var next_i:Int = curr[0]+dirs[j][0] 55 var next_j:Int = curr[1]+dirs[j][1] 56 if inBounds(forest, next_i , next_j) && !visited[next_i][next_j] 57 { 58 visited[next_i][next_j] = true 59 queue.append([next_i, next_j]) 60 } 61 } 62 } 63 level += 1 64 } 65 return -1 66 } 67 68 func inBounds(_ forest: [[Int]],_ i:Int,_ j:Int) -> Bool 69 { 70 return (i >= 0) && (i <= forest.count-1) 71 && (j >= 0) && (j <= forest[0].count - 1) && (forest[i][j] != 0) 72 } 73 }