LeetCode 542. 01 Matrix

原题链接在这里：https://leetcode.com/problems/01-matrix/description/

题目：

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1: 
Input:

0 0 0
0 1 0
0 0 0

Output:

0 0 0
0 1 0
0 0 0

Example 2: 
Input:

0 0 0
0 1 0
1 1 1

Output:

0 0 0
0 1 0
1 2 1

Note:

1. The number of elements of the given matrix will not exceed 10,000.
2. There are at least one 0 in the given matrix.
3. The cells are adjacent in only four directions: up, down, left and right.

题解：

先把非0点标记成Integer.MAX_VALUE. 然后从每个0点开始做BFS. 若是对周围的点更新出更小的值就把更小值放入queue中用作将来的BFS.

Time Complexity: O(m*n). m = matrix.length, n = matrix[0].length. 每个点最多进入queue 4 次.

Space: O(m*n). queue size.

AC Java:

class Solution {
    public int[][] updateMatrix(int[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return matrix;
        }
        
        int m = matrix.length;
        int n = matrix[0].length;
        
        LinkedList<int []> que = new LinkedList<int []>();
        for(int i = 0; i<m; i++){
            for(int j = 0; j<n; j++){
                if(matrix[i][j] == 0){
                    que.add(new int[]{i, j});
                }else{
                    matrix[i][j] = Integer.MAX_VALUE;
                }
            }
        }
        
        int [][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        
        while(!que.isEmpty()){
            int [] cur = que.poll();
            for(int [] dir : dirs){
                int dx = cur[0] + dir[0];
                int dy = cur[1] + dir[1];
                if(dx<0 || dx>=m || dy<0 || dy>=n || matrix[cur[0]][cur[1]]+1>=matrix[dx][dy]){
                    continue;
                }
                
                matrix[dx][dy] = matrix[cur[0]][cur[1]]+1;
                que.add(new int[]{dx, dy});
            }
        }
        
        return matrix;
    }
}