这道题折腾了我好久，dfs弄出来了但是时间复杂度太大了，重复搜索了太多的点，dfs代码（1252ms）:

class Solution {
public:
    void dfs(vector<vector<int>>& mat,int i,int j,int dis,int& ans){
        int m = mat.size();
        int n = mat[0].size();
        if(i<0||i>=m||j<0||j>=n){
            return ;
        }
        if(dis>ans) return ;
        if(mat[i][j]!=1){
            if(mat[i][j]==0) ans = min(ans,dis);
            return ;
        }
        mat[i][j] = 2;
        dfs(mat,i-1,j,dis+1,ans);
        dfs(mat,i+1,j,dis+1,ans);
        dfs(mat,i,j-1,dis+1,ans);
        dfs(mat,i,j+1,dis+1,ans);
        mat[i][j] = 1;
    }
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        int m = mat.size();
        int n = mat[0].size();
        vector<vector<int>> res(m,vector<int>(n,0));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(mat[i][j])   {
                    int ans = 1e9;
                    dfs(mat,i,j,0,ans);
                    res[i][j] = ans;
                }
            }
        }
        return res;
    }
};

题解用到的是BFS：
一开始写的dir在while循环内部，由于频繁申请空间，效率和dfs无异，改为in[][]或置到while外部204ms.

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
        queue<pair<int,int>> q;
        int m = mat.size();
        int n = mat[0].size();
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(mat[i][j]==0){
                    q.push(make_pair(i,j));
                }else{
                    mat[i][j] = m+n;
                }
            }
        }
        vector<vector<int>> dir = {{-1,0},{1,0},{0,1},{0,-1}};
        while(!q.empty()){
            pair<int,int> p = q.front();
            q.pop();
            for(int d=0;d<4;d++){
                int x =p.first,y = p.second;
                int r = x+dir[d][0];
                int c = y+dir[d][1];
                if(r<0||r>=m||c<0||c>=n||mat[r][c]<=mat[x][y]) continue;
                q.push(make_pair(r,c));
                mat[r][c]=mat[x][y]+1;
            }
        }
        return mat;
    }
};

最后我要放大招了，看了大神代码：O(m*n)184ms，里面那两个双重循环是怎么想到的啊QWQ

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> res(m, vector<int>(n, INT_MAX - 1));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) res[i][j] = 0;
                else {
                    if (i > 0) res[i][j] = min(res[i][j], res[i - 1][j] + 1);
                    if (j > 0) res[i][j] = min(res[i][j], res[i][j - 1] + 1);
                }
            }
        }
        
        for (int i = m -1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (res[i][j] != 0 && res[i][j] != 1) {
                    if (i < m - 1) res[i][j] = min(res[i][j], res[i + 1][j] + 1);
                    if (j < n - 1) res[i][j] = min(res[i][j], res[i][j + 1] + 1);
                }
            }
        }
        return res;
    }
};