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:

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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关键词 "...find the distance of the nearest 0 for each cell..."->最短路径问题->使用BFS:

1. 需要使用queue来实现BFS的一层一层遍历

2. 需要使用哈希表来避免重复遍历 =>不需要

对于图，需要找出点和边：

1.点：每个cell

2.边：每个cell和它上下左右四个cell是联通的

图需要有起点，这道题图的起点是所有是0的cell。

 

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        vector<vector<int>> dist(matrix.size(),vector<int>(matrix[0].size())); //store result
        queue<pair<int,int>> queue; //BFS 
        for(int i=0;i<matrix.size();i++){
            for(int j=0;j<matrix[0].size();j++){
                if(matrix[i][j]==0){
                    dist[i][j] = 0;     //root nodes
                    queue.push(make_pair(i,j));
                }else{
                    dist[i][j]=INT_MAX; //initialization
                }
            }
        }
        
        int dirs[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
        
        //BFS implementation
        while(!queue.empty()){
            pair<int,int> curr = queue.front(); //current cell
            queue.pop(); 
            
            //update its neighbors: 4 directions 
            for(auto dir: dirs){
                int x = dir[0] + curr.first;
                int y = dir[1] + curr.second;
                
                //check neighbor is within the grid
                if(x>=0&&y>=0&&x<matrix.size()&&y<matrix[0].size()){
                    //if the neighbor cell is 0-cell, still 0; otherwise, it can be arrived from current node=>value=current_vaule +1
                    int target = (matrix[x][y]==0?0:dist[curr.first][curr.second]+1); 

                    //no need for hashtable visited checking, if the node was visited before, dist[x][y] must be smaller than target
                    if(target<dist[x][y]){ 
                        dist[x][y] = target;
                        queue.push(make_pair(x,y));
                    }
                }
            }
        }
        
        return dist;
    }
};