1. 题目

给定一个整数 n, 返回从 1 到 n 的字典顺序。

例如，
给定 n =1 3，返回 [1,10,11,12,13,2,3,4,5,6,7,8,9] 。
请尽可能的优化算法的时间复杂度和空间复杂度。 输入的数据 n 小于等于 5,000,000。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/lexicographical-numbers
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

class Solution {
public:vector<int> lexicalOrder(int n) {vector<int> ans(n);	//n=1000int cur = 1;for(int i = 0; i < n; ++i){ans[i] = cur;if(cur*10 <= n)cur *= 10;			//1，10,100,1000else{if(cur >= n)		//1000cur /= 10;cur += 1;			//101。。。109while(cur%10 == 0)	//110cur /= 10;		//11}}return ans;}
};

n = 1000
[1,10,100,1000,101,102,103,104,105,106,107,108,109,
11,110,111,112,113,114,115,116,117,118,119,12,120,
121,122,123,124,125,126,127,128,129,13,130,131,132,
133,134,135,136,137,138,139,14,140,141,142,143,144,
145,146,147,148,149,15,150,151,152,153,154,155,156,
157,158,159,16,160,161,162,163,164,165,166,167,168,.

class Solution {vector<int> ans;
public:vector<int> lexicalOrder(int n) {for(int i=1; i<=9; i++)dfs(i, n);return ans;}void dfs(int cur, int n){if(cur > n)return;ans.push_back(cur);for(int i=0;i<10;i++){if(10*cur + i > n)return;dfs(10*cur+i, n);}}
};