目录

牛客_腐烂的苹果（多源 BFS）

题目解析

C++代码

Java代码

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牛客_腐烂的苹果（多源 BFS）

腐烂的苹果_牛客题霸_牛客网

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题目解析

多源 BFS 问题，固定套路，BFS学习 ：Offer必备算法28_多源BFS_四道力扣题详解（由易到难）_bfs算法题力扣-CSDN博客

C++代码

class Solution
{
    int m = 0, n = 0;
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    bool vis[1010][1010] = { 0 };
    public:
    int rotApple(vector<vector<int> >& grid)
    {
        m = grid.size(), n = grid[0].size();
        queue<pair<int, int>> q;
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                if(grid[i][j] == 2)
                    q.push({i, j});

        int ret = 0;
        while(q.size())
        {
            int sz = q.size();
            ret++;
            while(sz--)
            {
                auto [a, b] = q.front();
                q.pop();
                for(int i = 0; i < 4; i++)
                {
                    int x = a + dx[i], y = b + dy[i];
                    if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1
                       && !vis[x][y])
                    {
                        vis[x][y] = true;
                        q.push({x, y});
                    }
                }
            }
        }
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                if(grid[i][j] == 1 && !vis[i][j])
                    return -1;

        return ret - 1;
    }
};

Java代码

import java.util.*;
public class Solution
{
    int[] dx = {0, 0, 1, -1};
    int[] dy = {1, -1, 0, 0};
    public int rotApple (ArrayList<ArrayList<Integer>> grid)
    {
        int m = grid.size();
        int n = grid.get(0).size();
        boolean[][] vis = new boolean[m][n];
        Queue<int[]> q = new LinkedList<>();
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(grid.get(i).get(j) == 2)
                {
                    q.add(new int[]{i, j});
                }
            }
        }
        int ret = 0;
        while(!q.isEmpty())
        {
            int sz = q.size();
            while(sz-- != 0)
            {
                int[] t = q.poll();
                int a = t[0], b = t[1];
                for(int i = 0; i < 4; i++)
                {
                    int x = a + dx[i], y = b + dy[i];
                    if(x >= 0 && x < m && y >= 0 && y < n && vis[x][y] == false
                       && grid.get(x).get(y) == 1)
                    {
                        vis[x][y] = true;
                        q.add(new int[]{x, y});
                    }
                }
            }
            ret++;
        }
        // 判断剩余的苹果
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(grid.get(i).get(j) == 1 && !vis[i][j])
                {
                    return -1;
                }
            }
        }
        return ret - 1;
    }
}