542. 01 矩阵

给定一个由 0 和 1 组成的矩阵 mat ，请输出一个大小相同的矩阵，其中每一个格子是 mat 中对应位置元素到最近的 0 的距离。

两个相邻元素间的距离为 1 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/01-matrix
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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import java.util.LinkedList;
import java.util.Queue;

class Solution {

    private int[] dx = {0, 1, 0, -1};

    private int[] dy = {1, 0, -1, 0};

    public int[][] updateMatrix(int[][] mat) {
        if (mat == null || mat.length == 0 || mat[0].length == 0) {
            return new int[0][0];
        }

        int n = mat.length, m = mat[0].length;
        int[][] ret = new int[n][m];
        Queue<Operation> queue = new LinkedList<>();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (mat[i][j] == 0) {
                    queue.offer(new Operation(i, j, 0));
                }
            }
        }

        while (!queue.isEmpty()) {
            Operation operation = queue.poll();
            ret[operation.x][operation.y] = operation.step;
            for (int i = 0; i < 4; ++i) {
                int x = dx[i] + operation.x;
                int y = dy[i] + operation.y;
                if (x >= 0 && x < n && y >= 0 && y < m && mat[x][y] == 1) {
                    queue.offer(new Operation(x, y, operation.step + 1));
                    mat[x][y] = 0;
                }
            }
        }
        return ret;
    }
}

class Operation {
    int x;
    int y;
    int step;

    public Operation(int x, int y, int step) {
        this.x = x;
        this.y = y;
        this.step = step;
    }
}

动态规划

import java.util.LinkedList;
import java.util.Queue;

class Solution {

    private int[] dx = {0, 1, 0, -1};

    private int[] dy = {1, 0, -1, 0};

    public int[][] updateMatrix(int[][] mat) {
        if (mat == null || mat.length == 0 || mat[0].length == 0) {
            return new int[0][0];
        }

        int n = mat.length, m = mat[0].length;
        int[][] ret = new int[n][m];

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (mat[i][j] == 1) {
                    ret[i][j] = Integer.MAX_VALUE;
                }
            }
        }

        for (int i = 1; i < m; ++i) {
            if (mat[0][i] == 1 && ret[0][i - 1] != Integer.MAX_VALUE) {
                ret[0][i] = Math.min(ret[0][i], ret[0][i - 1] + 1);
            }
        }

        for (int i = 1; i < n; ++i) {
            if (mat[i][0] == 1 && ret[i - 1][0] != Integer.MAX_VALUE) {
                ret[i][0] = Math.min(ret[i][0], ret[i - 1][0] + 1);
            }
            for (int j = 1; j < m; ++j) {
                int previous = Math.min(ret[i][j - 1], ret[i - 1][j]);
                if (mat[i][j] == 1 && previous != Integer.MAX_VALUE) {
                    ret[i][j] = Math.min(ret[i][j], previous + 1);
                }
            }
        }

        for (int i = m - 2; i >= 0; --i) {
            if (mat[n - 1][i] == 1 && ret[n - 1][i + 1] != Integer.MAX_VALUE) {
                ret[n - 1][i] = Math.min(ret[n - 1][i], ret[n - 1][i + 1] + 1);
            }
        }

        for (int i = n - 2; i >= 0; --i) {
            if (mat[i][m - 1] == 1 && ret[i + 1][m - 1] != Integer.MAX_VALUE) {
                ret[i][m - 1] = Math.min(ret[i][m - 1], ret[i + 1][m - 1] + 1);
            }
            for (int j = m - 2; j >= 0; --j) {
                int previous = Math.min(ret[i + 1][j], ret[i][j + 1]);
                if (mat[i][j] == 1 && previous != Integer.MAX_VALUE) {
                    ret[i][j] = Math.min(ret[i][j], previous + 1);
                }
            }
        }

        return ret;
    }
}