386. Lexicographical Numbers

發表於 | 分類於

problem

solution

option 1 - iterative

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
class Solution {
public:
    vector<int> lexicalOrder(int n) {
        vector<int> ret(n);
        int cur = 1;
        for(int i=0;i<n;++i){
            ret[i] = cur;
            if(cur*10 <=n) cur*=10;
            else{
                if(cur>=n) cur/=10;
                cur++;
                while(cur%10==0) cur/=10;
            }
        }
        return ret;
    }
};

option 2 - dfs

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
class Solution {
public:
    void dfs(int cur, int n , vector<int> & ret){
        if(cur>n) return;
        ret.push_back(cur);
        for(int i=0;i<=9;++i){
            if(cur*10 +i <= n) dfs(cur*10+i,n,ret);
            else break;
        }
    }
    vector<int> lexicalOrder(int n) {
        vector<int> ret;
        for(int i=1;i<=9;++i) dfs(i,n,ret);
        return ret;
        
    }
};

analysis

  • option 1
    • time complexity O(n)
    • space complexity O(1)
  • option 2
    • time complexity O(n)
    • space complexity O(1)