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 00 1 00 0 0

Output:

0 0 00 1 00 0 0

Example 2:
Input:

0 0 00 1 01 1 1

Output:

0 0 00 1 01 2 1

Note:

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

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Solution:

Tips:

iterator matrix from left up corner to right down. then from right down to left up.

dp[i][j] = Math.min(dp[i][j], dp[i - 1][j] + 1, dp[i + 1][j] + 1, dp[i][j + 1] + 1, dp[i] [j - 1] + 1);

Java Code:

public class Solution {    public List<List<Integer>> updateMatrix(List<List<Integer>> matrix) {        List<List<Integer>> result = new ArrayList<>();        if (null == matrix) {            return result;        }                int m = matrix.size();        int n = matrix.get(0).size();        int maxDistance = m * n;                int[][] distance = new int[m][n];        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                distance[i][j] = matrix.get(i).get(j) == 0 ? 0 : maxDistance;            }        }                for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                distance[i][j] = i - 1 >= 0 ? Math.min(distance[i - 1][j] + 1, distance[i][j]) : distance[i][j];                distance[i][j] = j - 1 >= 0 ? Math.min(distance[i][j - 1] + 1, distance[i][j]) : distance[i][j];            }        }                for (int i = m - 1; i >= 0; i--) {            for (int j = n - 1; j >= 0; j--) {                distance[i][j] = i + 1 < m ? Math.min(distance[i + 1][j] + 1, distance[i][j]) : distance[i][j];                distance[i][j] = j + 1 < n ? Math.min(distance[i][j + 1] + 1, distance[i][j]) : distance[i][j];            }        }        for (int i = 0; i < m; i++) {            List<Integer> lst = new ArrayList<>();            for (int j = 0; j < n; j++) {                lst.add(distance[i][j]);            }            result.add(lst);        }                return result;    }}