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Formatted question description: https://leetcode.ca/all/994.html
994. Rotting Oranges (Medium)
In a given grid, each cell can have one of three values:
- the value
0representing an empty cell; - the value
1representing a fresh orange; - the value
2representing 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 <= grid.length <= 101 <= grid[0].length <= 10grid[i][j]is only0,1, or2.
Related Topics: Breadth-first Search
Similar Questions:
Solution 1. BFS
// OJ: https://leetcode.com/problems/rotting-oranges/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
int orangesRotting(vector<vector<int>>& A) {
int M = A.size(), N = A[0].size(), ans = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
queue<pair<int, int>> q;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == 2) q.emplace(i, j);
}
}
if (q.empty()) ans = 1;
while (q.size()) {
int cnt = q.size();
++ans;
while (cnt--) {
auto [x, y] = q.front();
q.pop();
for (auto &dir : dirs) {
int a = x + dir[0], b = y + dir[1];
if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 1) continue;
A[a][b] = 2;
q.emplace(a, b);
}
}
}
for (auto &row : A) {
for (int x : row) {
if (x == 1) return -1;
}
}
return ans - 1;
}
};
-
class Solution { public int orangesRotting(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) return 0; int rows = grid.length, columns = grid[0].length; final int BLOCK = -1; final int WHITE = 0; final int GRAY = 1; final int BLACK = 2; int[][] colors = new int[rows][columns]; int[][] distances = new int[rows][columns]; Queue<int[]> queue = new LinkedList<int[]>(); for (int i = 0; i < rows; i++) { for (int j = 0; j < columns; j++) { int value = grid[i][j]; if (value == 2) { distances[i][j] = 0; colors[i][j] = GRAY; queue.offer(new int[]{i, j}); } else if (value == 1) { distances[i][j] = Integer.MAX_VALUE; colors[i][j] = WHITE; } else { distances[i][j] = -1; colors[i][j] = BLOCK; } } } int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; while (!queue.isEmpty()) { int[] cell = queue.poll(); int row = cell[0], column = cell[1]; int distance = distances[row][column]; for (int[] direction : directions) { int newRow = row + direction[0], newColumn = column + direction[1]; if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns) { if (colors[newRow][newColumn] == WHITE) { distances[newRow][newColumn] = distance + 1; colors[newRow][newColumn] = GRAY; queue.offer(new int[]{newRow, newColumn}); } } } colors[row][column] = BLACK; } int time = 0; for (int i = 0; i < rows; i++) { for (int j = 0; j < columns; j++) { time = Math.max(time, distances[i][j]); if (time == Integer.MAX_VALUE) return -1; } } return time == Integer.MAX_VALUE ? -1 : time; } } ############ class Solution { public int orangesRotting(int[][] grid) { int m = grid.length, n = grid[0].length; int cnt = 0; Deque<int[]> q = new LinkedList<>(); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid[i][j] == 2) { q.offer(new int[] {i, j}); } else if (grid[i][j] == 1) { ++cnt; } } } int ans = 0; int[] dirs = {1, 0, -1, 0, 1}; while (!q.isEmpty() && cnt > 0) { ++ans; for (int i = q.size(); i > 0; --i) { int[] p = q.poll(); for (int j = 0; j < 4; ++j) { int x = p[0] + dirs[j]; int y = p[1] + dirs[j + 1]; if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) { grid[x][y] = 2; --cnt; q.offer(new int[] {x, y}); } } } } return cnt > 0 ? -1 : ans; } } -
// OJ: https://leetcode.com/problems/rotting-oranges/ // Time: O(MN) // Space: O(MN) class Solution { public: int orangesRotting(vector<vector<int>>& A) { queue<pair<int, int>> q; int M = A.size(), N = A[0].size(), step = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} }; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (A[i][j] == 2) q.emplace(i, j); } } while (q.size()) { for (int cnt = q.size(); cnt--;) { auto [x, y] = q.front(); q.pop(); for (auto &[dx, dy] : dirs) { int a = x + dx, b = y + dy; if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 1) continue; A[a][b] = 2; q.emplace(a, b); } } step++; } for (auto &row : A) { for (int x : row) { if (x == 1) return -1; } } return max(0, step - 1); } }; -
class Solution: def orangesRotting(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) q = deque() cnt = 0 for i in range(m): for j in range(n): if grid[i][j] == 2: q.append((i, j)) elif grid[i][j] == 1: cnt += 1 ans = 0 while q and cnt: ans += 1 for _ in range(len(q)): i, j = q.popleft() for a, b in [[0, 1], [0, -1], [1, 0], [-1, 0]]: x, y = i + a, j + b if 0 <= x < m and 0 <= y < n and grid[x][y] == 1: cnt -= 1 grid[x][y] = 2 q.append((x, y)) return ans if cnt == 0 else -1 ############ class Solution(object): def orangesRotting(self, grid): """ :type grid: List[List[int]] :rtype: int """ M, N = len(grid), len(grid[0]) fresh = 0 q = collections.deque() for i in range(M): for j in range(N): if grid[i][j] == 1: fresh += 1 elif grid[i][j] == 2: q.append((i, j)) if fresh == 0: return 0 dirs = [(0, 1), (0, -1), (-1, 0), (1, 0)] step = 0 while q: size = len(q) for i in range(size): x, y = q.popleft() for d in dirs: nx, ny = x + d[0], y + d[1] if nx < 0 or nx >= M or ny < 0 or ny >= N or grid[nx][ny] != 1: continue grid[nx][ny] = 2 q.append((nx, ny)) fresh -= 1 step += 1 if fresh != 0: return -1 return step - 1 -
func orangesRotting(grid [][]int) int { m, n := len(grid), len(grid[0]) cnt := 0 var q [][]int for i := 0; i < m; i++ { for j := 0; j < n; j++ { if grid[i][j] == 2 { q = append(q, []int{i, j}) } else if grid[i][j] == 1 { cnt++ } } } ans := 0 dirs := []int{-1, 0, 1, 0, -1} for len(q) > 0 && cnt > 0 { ans++ for i := len(q); i > 0; i-- { p := q[0] q = q[1:] for j := 0; j < 4; j++ { x, y := p[0]+dirs[j], p[1]+dirs[j+1] if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 { cnt-- grid[x][y] = 2 q = append(q, []int{x, y}) } } } } if cnt > 0 { return -1 } return ans } -
function orangesRotting(grid: number[][]): number { const m = grid.length; const n = grid[0].length; let count = 0; const queue = []; for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { if (grid[i][j] === 1) { count++; } else if (grid[i][j] === 2) { queue.push([i, j]); } } } let res = 0; const dris = [1, 0, -1, 0, 1]; while (count !== 0 && queue.length !== 0) { for (let i = queue.length; i > 0; i--) { const [x, y] = queue.shift(); for (let j = 0; j < 4; j++) { const newX = x + dris[j]; const newY = y + dris[j + 1]; if ( newX >= 0 && newX < m && newY >= 0 && newY <= n && grid[newX][newY] === 1 ) { grid[newX][newY] = 2; queue.push([newX, newY]); count--; } } } res++; } if (count != 0) { return -1; } return res; } -
use std::collections::VecDeque; impl Solution { pub fn oranges_rotting(mut grid: Vec<Vec<i32>>) -> i32 { let mut queue = VecDeque::new(); let m = grid.len(); let n = grid[0].len(); // 新鲜橘子数量 let mut count = 0; for i in 0..m { for j in 0..n { match grid[i][j] { 1 => count += 1, 2 => queue.push_back([i as i32, j as i32]), _ => (), } } } let mut res = 0; let dirs = [1, 0, -1, 0, 1]; while count != 0 && queue.len() != 0 { let mut len = queue.len(); while len != 0 { let [x, y] = queue.pop_front().unwrap(); for i in 0..4 { let new_x = x + dirs[i]; let new_y = y + dirs[i + 1]; if new_x >= 0 && new_x < m as i32 && new_y >= 0 && new_y < n as i32 && grid[new_x as usize][new_y as usize] == 1 { grid[new_x as usize][new_y as usize] = 2; queue.push_back([new_x, new_y]); count -= 1; } } len -= 1; } res += 1; } if count != 0 { return -1; } res } }