Rotting Oranges

Description

You are given an m x n grid where each cell can have one of three values:

• 0 representing an empty cell,
• 1 representing a fresh orange, or
• 2 representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent 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.

Example 1:

Input: grid = [[2,1,1],[1,1,0],[0,1,1]] Output: 4

Example 2:

Input: grid = [[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: grid = 0,2 Output: 0 Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10
• grid[i][j] is 0, 1, or 2.

Code

similar to Flood Fill，因為要求 minimum minutes，因此要使用 BFS，要思考的點在於，若有很多個 rotten oranges，要讓他們同時開始感染其他健康的 oranges，因為 BFS 是一個 level 一個 level 往下做，因此所有 initial rotten oranges 都會在 level 0。

注意 vector<int> dir = {-1, 0, 1, 0, -1};的使用。

注意一開始時設 ans = -1 。

class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        queue<pair<int, int>> q;
        int fresh = 0;
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == 2) q.push({i, j});
                if(grid[i][j] == 1) fresh++;
            }
        }
 
        vector<int> dir = {-1, 0, 1, 0, -1};
        int ans = -1;
        while(!q.empty()) {
            int s = q.size();
            while(s--) {
                auto node = q.front();
                q.pop();
                for(int i = 0; i < 4; i++) {
                    int x = node.first + dir[i];
                    int y = node.second + dir[i+1];
                    if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                        grid[x][y] = 2;
                        fresh--;
                        q.push({x, y});
                    }
                }
            }
 
            ans++;
        }
 
        if(fresh) return -1;
        if(ans == -1) return 0;
        return ans;
    }
 
};

Source

• Rotting Oranges - LeetCode