[LeetCode] 994. Rotting Oranges 腐爛的橘子

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.

這道題說給的一個 mxn 大小的格子上有些新鮮和腐爛的橘子，每一分鍾腐爛的橘子都會傳染給其周圍四個中的新鮮橘子，使得其也變得腐爛。現在問需要多少分鍾可以使得所有的新鮮橘子都變腐爛，無法做到時返回 -1。由於這里新鮮的橘子自己不會變腐爛，只有被周圍的腐爛橘子傳染才會，所以當新鮮橘子周圍不會出現腐爛橘子的時候，那么這個新鮮橘子就不會腐爛，這才會有返回 -1 的情況。這道題就是個典型的廣度優先遍歷 Breadth First Search，並沒有什么太大的難度，先遍歷一遍整個二維數組，統計出所有新鮮橘子的個數，並把腐爛的橘子坐標放入一個隊列 queue，之后進行 while 循環，循環條件是隊列不會空，且 freshLeft 大於0，使用層序遍歷的方法，用個 for 循環在內部。每次取出隊首元素，遍歷其周圍四個位置，越界或者不是新鮮橘子都跳過，否則將新鮮橘子標記為腐爛，加入隊列中，並且 freshLeft 自減1。每層遍歷完成之后，結果 res 自增1，最后返回的時候，若還有新鮮橘子，即 freshLeft 大於0時，返回 -1，否則返回 res 即可，參見代碼如下：

class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int res = 0, m = grid.size(), n = grid[0].size(), freshLeft = 0;
        queue<vector<int>> q;
        vector<vector<int>> dirs{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) ++freshLeft;
                else if (grid[i][j] == 2) q.push({i, j});
            }
        }
        while (!q.empty() && freshLeft > 0) {
            for (int i = q.size(); i > 0; --i) {
                auto cur = q.front(); q.pop();
                for (int k = 0; k < 4; ++k) {
                    int x = cur[0] + dirs[k][0], y = cur[1] + dirs[k][1];
                    if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] != 1) continue;
                    grid[x][y] = 2;
                    q.push({x, y});
                    --freshLeft;
                }
            }
            ++res;
        }
        return freshLeft > 0 ? -1 : res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/994

類似題目：

Walls and Gates

參考資料：

https://leetcode.com/problems/rotting-oranges/

https://leetcode.com/problems/rotting-oranges/discuss/238681/Java-Clean-BFS-Solution-with-comments

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