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

Solution：BFS 染色

思路：
我們可以首先遍歷一次矩陣，將值為0的點都存入queue，將值為1的點改為INT_MAX。
然后開始BFS遍歷，從queue中取出一個數(shù)字，遍歷其周圍四個點，如果越界或者周圍點的值小于等于當前值，則直接跳過。因為周圍點的距離更小的話(已被更好的染過)，就沒有更新的必要，否則將周圍點的值更新為當前值加1，然后把周圍點的坐標加入queue。
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

public class Solution {
    public int[][] updateMatrix(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        
        Queue<int[]> queue = new LinkedList<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) {
                    queue.offer(new int[] {i, j});
                }
                else {
                    matrix[i][j] = Integer.MAX_VALUE;
                }
            }
        }
        
        int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            for (int[] d : dirs) {
                int r = cell[0] + d[0];
                int c = cell[1] + d[1];
                if (r < 0 || r >= m || c < 0 || c >= n || 
                    matrix[r][c] <= matrix[cell[0]][cell[1]] + 1) continue;
                queue.add(new int[] {r, c});
                matrix[r][c] = matrix[cell[0]][cell[1]] + 1;
            }
        }
        
        return matrix;
    }
}