【BFS】腐烂的橘子

https://leetcode.cn/problems/rotting-oranges/description/?envType=study-plan-v2&envId=top-100-liked

1. 从一个点向上下左右移动并且判断是否边界可以用

for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
  nx = x + dx
  ny = y + dy
  if 0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] == 1:

2. BFS常规技巧：多源广度优先 = 同层节点遍历 = 先获取队列size，然后弹出size个元素，就可以恰好遍历完这个批次的队列（哎呀好久没做题都忘光了XD，连这个都要写）

size = len(queue)
for _ in range(size):
    x, y = queue.popleft()

class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        cols = len(grid[0])

        queue = deque()
        fresh_oranges = 0

        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 2:
                    queue.append((i, j))
                elif grid[i][j] == 1:
                    fresh_oranges += 1

       minutes = 0
        if fresh_oranges == 0:
            return 0

        while queue:
            size = len(queue)

            for _ in range(size):
                x, y = queue.popleft()

                for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
                    nx = x + dx
                    ny = y + dy

                    if 0 <= nx < rows and 0 <= ny < cols and grid[nx][ny] == 1:
                        grid[nx][ny] = 2
                        queue.append((nx, ny))
                        fresh_oranges -= 1
            if queue:
                minutes += 1


        return -1 if fresh_oranges != 0 else minutes