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 0
0 1 0
0 0 0
Output:
0 0 0
0 1 0
0 0 0

Example 2: 
Input:
0 0 0
0 1 0
1 1 1
Output:
0 0 0
0 1 0
1 2 1

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

难度：medium

题目：
给定由0､1组成的矩阵，算出每个元素离0最近的距离。
相离的两个单元格距离为1.

注意：
给出的矩阵元素不超过10,000.
矩阵至少包含一个0.
毗邻的定义为上下左右4个方向。

思路：
从左上到右下遍历计算一遍，然后再自右下到左上遍历计算一遍。

Runtime: 19 ms, faster than 64.23% of Java online submissions for 01 Matrix.

public class Solution {
    public int[][] updateMatrix(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int maxDis = m * n;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] > 0) {
                    int upDis = (0 == i) ? maxDis : matrix[i - 1][j] + 1;
                    int leftDis = (0 == j) ? maxDis : matrix[i][j - 1] + 1;
                    matrix[i][j] = Math.min(upDis, leftDis);
                }
            }
        }
        
        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (matrix[i][j] > 0) {
                    int downDis = (m - 1 == i) ? maxDis : matrix[i + 1][j] + 1;
                    int rightDis = (n - 1 == j) ? maxDis : matrix[i][j + 1] + 1;
                    matrix[i][j] = Math.min(matrix[i][j], Math.min(downDis, rightDis));
                }
            }
        }

        return matrix;
    }
}