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;
}
};