542. 01 Matrix

问题：

给定一个由 0 和 1 构成的二维数组。

求每个cell到最近0的距离。

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.

　　

解法：BFS

思想：将所有 0 的cell加入queue

然后向四周蔓延，遍历完整个数组。

每蔓延一层，距离+1

层数即是queue的遍历level。

 

代码参考：

class Solution {
public:
    int m,n;
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        n=matrix.size();
        if(n==0) return matrix;
        m=matrix[0].size();
        vector<int> dir={1,0,-1,0,1};
        vector<vector<int>> res(n, vector<int>(m,0));
        vector<vector<bool>> visited(n, vector<bool>(m,false));
        queue<pair<int, int>> q;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(matrix[i][j]==0) {
                    q.push({i,j});
                    visited[i][j]=true;
                }
            }
        }
        int cur_i, cur_j;
        int x, y;
        int level = 0;
        while(!q.empty()) {
            int sz = q.size();
            level++;
            for(int i=0; i<sz; i++) {
                cur_i = q.front().first;
                cur_j = q.front().second;
                q.pop();
                for(int j=1; j<5; j++) {
                    x=cur_i+dir[j-1];
                    y=cur_j+dir[j];
                    if(x<0 || y<0 || x>=n || y>=m || visited[x][y]) continue;
                    res[x][y]=level;
                    q.push({x,y});
                    visited[x][y] = true;
                }
            }
        }
        return res;
    }
};