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.

这道题DFS的表现不好，一直提示error: java.lang.StackOverflowError。DP是最快的，先从左上向右下遍历一遍，在从右下向左上遍历一遍，得到结果。代码如下：

public class Solution {    public int[][] updateMatrix(int[][] matrix) {        if (matrix == null || matrix.length == 0) {            return new int[0][0];        }        int m = matrix.length;        int n = matrix[0].length;        int[][] dp = new int[m][n];        int range = 10001;        for (int i = 0; i < m; i ++) {            for (int j = 0; j < n; j ++) {                if (matrix[i][j] == 0) {                    dp[i][j] = 0;                } else {                    int upCell = i > 0? dp[i - 1][j]: range;                    int leftCell = j > 0? dp[i][j - 1]: range;                    dp[i][j] = Math.min(upCell, leftCell) + 1;                }            }        }        for (int i = m - 1; i >= 0; i --) {            for (int j = n - 1; j >= 0; j --) {                if (matrix[i][j] == 0) {                    dp[i][j] = 0;                } else {                    int downCell = i < m - 1? dp[i + 1][j]: range;                    int rightCell = j < n - 1? dp[i][j + 1]: range;                    dp[i][j] = Math.min(dp[i][j], Math.min(downCell, rightCell) + 1);                }            }        }        return dp;    }}

BFS也可以解题，用第一遍遍历时，queue保存所有0的位置，所有的1设置为max_value。第二遍遍历时，如果上下左右的方向上有比此点大的点，那个点的值设置成此点值+1，存下那个点的坐标，一直遍历到queue为空。代码如下：

public class Solution {    public int[][] updateMatrix(int[][] matrix) {        int m = matrix.length;        int n = matrix[0].length;                Queue<int[]> queue = new LinkedList<>();        for (int i = 0; i < m; i++) {            for (int j = 0; j < n; j++) {                if (matrix[i][j] == 0) {                    queue.offer(new int[] {i, j});                }                else {                    matrix[i][j] = Integer.MAX_VALUE;                }            }        }                int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};                while (!queue.isEmpty()) {            int[] cell = queue.poll();            for (int[] d : dirs) {                int r = cell[0] + d[0];                int c = cell[1] + d[1];                if (r < 0 || r >= m || c < 0 || c >= n ||                     matrix[r][c] <= matrix[cell[0]][cell[1]] + 1) continue;                queue.add(new int[] {r, c});                matrix[r][c] = matrix[cell[0]][cell[1]] + 1;            }        }                return matrix;    }}