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.

思路：BFS错不了，但是一个个点做的话会TLE，想想也是，不同的节点的BFS有一定的重叠，其实完全可以一起BFS遍历啊！

还有值得注意的一点是，给的输入是List，一般都是int[]，所以这里就直接用输入，

package l542;import java.util.LinkedList;import java.util.List;import java.util.Queue;/* * do bfs at same time * 直接在输入的list对象上更改就好了 */public class Solution {    public List<List<Integer>> updateMatrix(List<List<Integer>> matrix) {    Queue<int[]> q = new LinkedList<int[]>();        for(int i=0; i<matrix.size(); i++) {    for(int j=0; j<matrix.get(0).size(); j++) {    if(matrix.get(i).get(j) == 0) {    q.add(new int[]{i, j});    } else {    matrix.get(i).set(j, -1);//用来标识是否找到了该点的最短值    }    }    }        while(!q.isEmpty()) {    int[] cell = q.remove();    int x = cell[0], y = cell[1];int step = matrix.get(x).get(y);    if(x > 0 && matrix.get(x-1).get(y) == -1){q.add(new int[]{x-1, y});matrix.get(x-1).set(y, step+1);}if(y > 0 && matrix.get(x).get(y-1) == -1) {q.add(new int[]{x, y-1});matrix.get(x).set(y-1, step+1);}if(x < matrix.size()-1 && matrix.get(x+1).get(y) == -1){q.add(new int[]{x+1, y});matrix.get(x+1).set(y, step+1);}if(y < matrix.get(0).size()-1 && matrix.get(x).get(y+1) == -1){q.add(new int[]{x, y+1});matrix.get(x).set(y+1, step+1);}    }        return matrix;    }}