Rotting Oranges

Description

In a given grid, each cell can have one of three values:

• the value 0 representing an empty cell;
• the value 1 representing a fresh orange;
• the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange.  If this is impossible, return -1 instead.

 

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: -1
Explanation:  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: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

 

Note:

1. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.

Solution(javascript)

/** 这题可以用 BFS
 * @param {number[][]} grid
 * @return {number}
 */
const orangesRotting = function (grid) {
  let freshCount = 0
  let count = -1
  let current = []
  for (let i = 0; i < grid.length; i++) {
    for (let j = 0; j < grid[i].length; j++) {
      if (grid[i][j] === 1) {
        freshCount += 1
      }
      if (grid[i][j] === 2) {
        current.push([i, j])
      }
    }
  }
  if (!freshCount) {
    return 0
  }
  while (current.length > 0) {
    count += 1
    const next = []
    const rotten = (i, j) => { // eslint-disable-line
      grid[i][j] = 2
      next.push([i, j])
      freshCount -= 1
    }
    for (const [i, j] of current) {
      if (grid[i - 1] && grid[i - 1][j] === 1) {
        rotten(i - 1, j)
      }
      if (grid[i + 1] && grid[i + 1][j] === 1) {
        rotten(i + 1, j)
      }
      if (grid[i][j - 1] === 1) {
        rotten(i, j - 1)
      }
      if (grid[i][j + 1] === 1) {
        rotten(i, j + 1)
      }
    }
    current = next
  }
  return freshCount === 0 ? count : -1
}