标签:二维数组 direct cto rect ado 维数 个数 queue 空间
腐烂的橘子。题意是给一个二维数组,用几个数字分别表示橘子的腐烂情况。每过去一分钟,任何腐烂的橘子的上下左右四个方向上的新鲜橘子都会腐烂。请求出全部橘子腐烂需要的分钟数,如果不会全都腐烂,则返回-1。
0
代表空单元格;
1
代表新鲜橘子;
2
代表腐烂的橘子。
例子,
Example 1:Input: [[2,1,1],[1,1,0],[0,1,1]]Output: 4
Example 2:Input: [[2,1,1],[0,1,1],[1,0,1]]Output: -1Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
Example 3:Input: [[0,2]]Output: 0Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
这是一个BFS的基本题。需要用queue去记录每个cell四周的四个cell在被更新过之后的腐烂情况。
时间O(mn)
空间O(mn)
1 /** 2 * @param {number[][]} grid 3 * @return {number} 4 */ 5 var orangesRotting = function (grid) { 6 let q = []; 7 let newFresh = 0; 8 let minutes = 0; 9 // push rotten oranges to the stack and count fresh oranges 10 for (let i = 0; i < grid.length; i++) { 11 for (let j = 0; j < grid[i].length; j++) { 12 if (grid[i][j] === 2) q.push([i, j]); 13 if (grid[i][j] === 1) newFresh++; 14 } 15 } 16 17 while (q.length && newFresh) { 18 let newQ = []; // queue for next minute 19 while (q.length) { 20 let badOrange = q.shift(); 21 let newRottens = infectOthers(grid, badOrange[0], badOrange[1], newQ); 22 newFresh -= newRottens; 23 } 24 minutes++; 25 q = newQ; 26 } 27 if (newFresh !== 0) return -1; 28 return minutes; 29 }; 30 31 // Infect surrounding oranges 32 // Return the number of newly infected oranges 33 var infectOthers = function (grid, i, j, q) { 34 let infected = 0; 35 if (i > 0 && grid[i - 1][j] === 1) { 36 grid[i - 1][j]++; 37 infected++; 38 q.push([i - 1, j]); 39 } 40 if (j > 0 && grid[i][j - 1] === 1) { 41 grid[i][j - 1]++; 42 infected++; 43 q.push([i, j - 1]); 44 } 45 if (i < grid.length - 1 && grid[i + 1][j] === 1) { 46 grid[i + 1][j]++; 47 infected++; 48 q.push([i + 1, j]); 49 } 50 if (j < grid[0].length - 1 && grid[i][j + 1] === 1) { 51 grid[i][j + 1]++; 52 infected++; 53 q.push([i, j + 1]); 54 } 55 return infected; 56 }
[LeetCode] 994. Rotting Oranges
标签:二维数组 direct cto rect ado 维数 个数 queue 空间
原文地址:https://www.cnblogs.com/aaronliu1991/p/12408644.html