[leetcode]386. Lexicographical Numbers

Analysis

normal day—— [每天刷題並不難0.0]

Given an integer n, return 1 - n in lexicographical order.
For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].
Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.

Implement

試圖直接暴力解決，意料之中的MLE T.T

class Solution {
public:
    vector<int> lexicalOrder(int n) {
        vector<string> str;
        for(int i=1; i<=n; i++)
            str.push_back(to_string(i));
        sort(str.begin(), str.end());
        vector<int> res;
        for(auto s:
 
str)
            res.push_back(stoi(s));
        return res;
    }
};

看了討論區的解答，發現這道題的關鍵其實是根據當前數字找到下一個數字。

class Solution {
public:
    vector<int> lexicalOrder(int n) {
        vector<int> res;
        int tmp = 1;
        for(int i=1; i<=n; i++){
            res.push_back(tmp);
            if
 
(tmp*10 <= n)
                tmp *= 10;
            else if(tmp%10 != 9 && tmp+1 <= n)
                tmp += 1;
            else{
                while(tmp%10 == 9 || tmp == n)
                    tmp /= 10;
                tmp += 1;
            }
        }
        return res;
    }
};

用DFS解決好像跟簡單易懂一些，可以把結果看成一系列的樹：

class Solution {
public:
    vector<int> lexicalOrder(int n) {
        for(int i=1; i<10; i++)
            dfs(i, n);
        return res;
    }
    void dfs(int cur, int n){
        if(cur > n)
            return ;
        res.push_back(cur);
        for(int i=0; i<10; i++){
            if(cur*10+i > n)
                return ;
            dfs(cur*10+i, n);
        }
    }
private:
    vector<int> res;
};