Sword finger offer II 107 Distance in matrix

The finger of the sword Offer II 107. The distance in the matrix 【 Medium question 】

Ideas ：

use BFS Sequence traversal , Find all the elements 0, From element 0 Start sequence traversal in four directions , set up ans The array stores the latest 0 distance , be ans in ,mat Matrix elements 0 The position value is 0, Encountered during traversal of each layer 1 Will ans Value is updated to element 0 Of ans value +1

Code ：

class Solution {
    
    
    static int[][] ans;
    static int[][] dir = {
    {
    0,-1},{
    0,1},{
    -1,0},{
    1,0}};
    static int m,n;
    public static int[][] updateMatrix(int[][] mat) {
    
        m = mat.length;
        n = mat[0].length;
        // Mark mat matrix (i,j) Whether the location element has been searched 
        boolean[][] seen = new boolean[m][n];
        // Storage mat In the matrix 0 The position coordinates of 
        Queue<int[]> queue = new LinkedList<>();
        // Storage mat matrix (i,j) The nearest location element 0 Distance of 
        ans = new int[m][n];
        for (int i = 0; i < m; i++) {
    
            for (int j = 0; j < n; j++) {
    
                if (mat[i][j] == 0){
    
                    // take mat The elements in the matrix 0 The position coordinates of are pressed into the queue 
                    queue.offer(new int[]{
    i,j});
                    // The current location of the tag has been searched 
                    seen[i][j] = true;
                }
            }
        }
        //BFS Traverse , When the element in the stack is empty , The loop ends 
        while(!queue.isEmpty()){
    
            // Take out the top element of the queue 0 The location of 
            int[] cell = queue.poll();
            // Take out the elements 0 Coordinates of 
            int i = cell[0],j = cell[1];
            for (int k = 0; k < 4; k++) {
    
                // The element 0 Search up, down, left and right , The location coordinates searched are (x,y)
                int x = i + dir[k][0],y = j + dir[k][1];
                // If (x,y) Does not exceed the matrix boundary and (x,y) The location has not been searched 
                if (x >= 0 && x < m && y >= 0 && y < n && !seen[x][y]){
    
                    //ans Of (x,y) Position value   be equal to  ans Of (i,j) Position value  + 1
                    ans[x][y] = ans[i][j] + 1;
                    // take (x,y) Position push into queue 
                    queue.offer(new int[]{
    x,y});
                    // take (x,y) The location is marked as searched 
                    seen[x][y] = true;
                }
            }
        }
        return ans;
    }
}