LeetCode 994. Rotting Oranges (Java版; Easy)

た入场券 2023-07-11 06:24 140阅读 0赞

### [welcome to my blog][] ###

## LeetCode 994. Rotting Oranges (Java版; Easy) ##

#### 题目描述 ####

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:

![format_png][]

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

#### 第一次做; 核心: 1)广度优先遍历, 使用队列+循环实现 2)队列中存放烂橘子 3) 当最后一个橘子被感染时, 该橘子入队, 但是在下一轮循环中不能感染任何橘子了, 也就是在下一轮循环中不会向队列中添加新元素; 所以在下一轮循环中不能执行res++操作了, 这就要求更新res的条件不仅仅是countDown==0还得满足队列不为空 ####

//想到一道题, 原地修改矩阵的值.....hot100或者top interview中的
    //BFS
    class Solution { 
        public int orangesRotting(int[][] grid) { 
            int n = grid.length, m = grid[0].length;
            //队列中存放烂橘子
            Queue<Cor> queue = new LinkedList<>();
            //找到所有腐烂的橘子, 入队
            for(int i=0; i<n; i++){ 
                for(int j=0; j<m; j++){ 
                    if(grid[i][j]==2){ 
                        queue.add(new Cor(i,j));
                    }
                }
            }
            //没有腐烂的橘子, 也就是队列为空, 此时如果grid中有1, 则返回-1, 如果没有1, 则返回0; 在代码最后统一处理
            // if(queue.isEmpty())
            // return 0;
    
            int res = 0;
            //烂橘子的个数, 初始时是队列中元素的个数
            int count = queue.size(), countDown;
            //bfs
            while(!queue.isEmpty()){ 
                countDown = count;
                count = 0;
                while(countDown>0){ 
                    Cor cur = queue.poll();
                    int x,y;
                    for(int[] arr : move){ 
                        x = cur.x + arr[0];
                        y = cur.y + arr[1];
                        if(valid(grid, x, y) && grid[x][y]==1){ 
                            count++;
                            grid[x][y]=2;
                            queue.add(new Cor(x, y));
                        }
                    }
                    countDown--;
                    //核心: !queue.isEmpty(); 感染到最后一个节点时, 这个节点在下一轮循环中已经不能感染其他节点了, 也就是队列为空, 此时不再增加res
                    if(countDown==0 && !queue.isEmpty()){ 
                        res++;
                    }
                }
            }
            for(int[] arr : grid){ 
                for(int a : arr){ 
                    if(a==1)
                        return -1;
                }
            }
            return res;
        }
        
        class Cor{ 
            int x;
            int y;
            Cor(int x, int y){ 
                this.x=x;
                this.y=y;
            }
        }
        
        private int[][] move = { { -1,0}, { 1,0}, { 0,-1},{ 0,1}};
        private boolean valid(int[][] grid, int i, int j){ 
            return i>=0 && i<grid.length && j>=0 && j<grid[0].length;
        }
    }

#### LeetCode最优解; 1)队列的元素是int\[\], 不用单独创建类 2) 用一个变量记录新鲜橘子的数量, 可以起到两个重要作用, 一个是如果第一次遍历grid时,如果没有新鲜橘子则直接返回0; 另一个是最后返回时, 如果新鲜橘子数量不为0则返回-1; 3)没有单独处理最后一个感染的橘子, 会导致count比正确的值大一, 处理方法: 返回count-1即可 ####

class Solution { 
        public int orangesRotting(int[][] grid) { 
            if(grid == null || grid.length == 0) return 0;
            int rows = grid.length;
            int cols = grid[0].length;
            Queue<int[]> queue = new LinkedList<>();
            int count_fresh = 0;
            //Put the position of all rotten oranges in queue
            //count the number of fresh oranges
            for(int i = 0 ; i < rows ; i++) { 
                for(int j = 0 ; j < cols ; j++) { 
                    if(grid[i][j] == 2) { 
                        queue.offer(new int[]{ i , j});
                    }
                    else if(grid[i][j] == 1) { 
                        count_fresh++;
                    }
                }
            }
            //if count of fresh oranges is zero --> return 0 
            if(count_fresh == 0) return 0;
            int count = 0;
            int[][] dirs = { { 1,0},{ -1,0},{ 0,1},{ 0,-1}};
            //bfs starting from initially rotten oranges
            while(!queue.isEmpty()) { 
                ++count;
                int size = queue.size();
                for(int i = 0 ; i < size ; i++) { 
                    int[] point = queue.poll();
                    for(int dir[] : dirs) { 
                        int x = point[0] + dir[0];
                        int y = point[1] + dir[1];
                        //if x or y is out of bound
                        //or the orange at (x , y) is already rotten
                        //or the cell at (x , y) is empty
                            //we do nothing
                        if(x < 0 || y < 0 || x >= rows || y >= cols || grid[x][y] == 0 || grid[x][y] == 2) continue;
                        //mark the orange at (x , y) as rotten
                        grid[x][y] = 2;
                        //put the new rotten orange at (x , y) in queue
                        queue.offer(new int[]{ x , y});
                        //decrease the count of fresh oranges by 1
                        count_fresh--;
                    }
                }
            }
            return count_fresh == 0 ? count-1 : -1;
        }
    }

[welcome to my blog]: https://littlehaes.com
[format_png]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9ub3RlLnlvdWRhby5jb20veXdzL3B1YmxpYy9yZXNvdXJjZS80MzkyNGVmYjU1NTg1NDYzMWQ3NDUxMTU4YmM5YTIzYi94bWxub3RlLzkxOTM0OTRFRTc2QTQ4MDQ5OENFMUMzOENDMEM3RjRDLzY0Njc5?x-oss-process=image/format,png