算法-二维矩阵各点到0值的距离

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算法之二维矩阵各点到0值的距离

  • 给定一个矩阵，返回各个节点到0点位置的最短路径

如:
{0,1,1,1}
{1,1,1,0}
{1,0,1,1}

返回:
[0, 1, 2, 1]
[1, 1, 1, 0]
[1, 0, 1, 1]

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package tmp;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
/**
 * Created by cdx0312
 */
public class Test {
    public static void main(String[] args) {
        int[][] arr = new int[3][4];
        arr[0] = new int[]{0,1,1,1};
        arr[1] = new int[]{1,1,1,0};
        arr[2] = new int[]{1,0,1,1};
        int[][] res = getResult(arr);
        for (int i = 0; i < 3; i++)
            System.out.println(Arrays.toString(res[i]));
    }
    private static int[][] getResult(int[][] arr) {
        int M = arr.length;
        int N = arr[0].length;
        Queue<Point> queue = new LinkedList<>();
        for (int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                if (arr[i][j] == 0)
                    queue.offer(new Point(i, j));
                else
                    arr[i][j] = Integer.MAX_VALUE;
            }
        }
        while (!queue.isEmpty()) {
            Point point = queue.poll();
            int x = point.x, y = point.y;
            int target = arr[x][y];
            if (x >= 1 && arr[x-1][y] > target + 1) {
                arr[x - 1][y] = arr[x][y] + 1;
                queue.offer(new Point(x-1, y));
            }
            if (x < M - 1 && arr[x+1][y] > target + 1) {
                arr[x + 1][y] = arr[x][y] + 1;
                queue.offer(new Point(x+1, y));
            }
            if (y >= 1 && arr[x][y-1] > target + 1) {
                arr[x][y - 1] = arr[x][y] + 1;
                queue.offer(new Point(x, y-1));
            }
            if (y < N - 1 && arr[x][y+1] > target + 1) {
                arr[x][y + 1] = arr[x][y] + 1;
                queue.offer(new Point(x, y+1));
            }
        }
        return arr;
    }
    static class Point {
        int x;
        int y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
}
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