題目描述
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 <= grid.length <= 10
1 <= grid[0].length <= 10
grid[i][j] is only 0, 1, or 2.
思路
bfs。
代碼
class Solution {
public:
vector<int> dirs {1, 0, -1, 0, 1};
int orangesRotting(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<int>> vis(n, vector<int>(m, 0));
queue<pair<int, int>> que;
int tot = 0;
for (int i=0; i<n; ++i) {
for (int j=0; j<m; ++j) {
if (grid[i][j] == 2) {
//vis[i][j] = 1;
que.push({i, j});
}
if (grid[i][j] == 1) tot++;
}
}
int step = 0;
while(!que.empty() && tot) {
step++;
int size = que.size();
while(size--) {
auto cur = que.front();
que.pop();
int x = cur.first;
int y = cur.second;
for (int i=0; i<4; ++i) {
int nx = x + dirs[i];
int ny = y + dirs[i+1];
if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
if (grid[nx][ny] == 0 || grid[nx][ny] == 2) continue;
grid[nx][ny] = 2;
tot--;
que.push({nx, ny});
}
}
}
return tot == 0 ? step : -1;
}
};
日期
2020 年 6 月 8 日 —— 明早要不要起來跑步。