542. 01 Matrix

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.

class Solution {
    public int[][] updateMatrix(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        Queue<int[]> q = new LinkedList<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0)
                    q.add(new int[]{i, j}); //把0元素加入队列中，以备波及影响周围元素
                else
                    matrix[i][j] = Integer.MAX_VALUE;//设为最大值，方便求0元素影响值
            }
        }
        //溅起的涟漪，代表传播的四个方向
        int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        while (!q.isEmpty()) {
            int[] cell = q.poll();
            for (int[] dirc : dirs) {
                //需要比较的下一位置元素
                int r = cell[0] + dirc[0];
                int c = cell[1] + dirc[1];

                //如果下一位置的元素值，小于等于（当前位置元素值+1），直接跳过
                if (r < 0 || r >= m || c < 0 || c >= n || matrix[r][c] <= (matrix[cell[0]][cell[1]] + 1)) 
                    continue;
                else {
                    matrix[r][c] = matrix[cell[0]][cell[1]] + 1;
                    q.offer(new int[]{r, c});
                }
            }
        }
        return matrix;        
    }
}