386. Lexicographical Numbers

Problem

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

.

Code

class Solution {
public:
vector<int> lexicalOrder(int n) {
vector<int> res;
if(n <= 9) {
for(int i = 1; i <= n; ++i)
res.push_back(i);
} else {
//如果大于9的时候就按照顺序 递归枚举
for(int i = 1; i <= 9; ++i) {
res.push_back(i);
dfs(res, i, n);
}
}
return res;
}

void dfs(vector<int>& res, int start, int n) {
int next_num = 10 * start;
if(next_num <= n) {
//计算差值 ，如果差值超过9那么 就从1-9继续下一层的搜索
int diff_num = n - next_num;
diff_num = diff_num > 9 ? 9 : diff_num;
for(int i = 0; i <= diff_num; ++i) {
int tmp_num = next_num + i;
res.push_back(tmp_num);
dfs(res, tmp_num, n);
}
}
}
};