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386. Lexicographical Numbers 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> lexicalOrder(int n) {
    vector<int> ans;
    int curr = 1;

    while (ans.size() < n) {
      ans.push_back(curr);
      if (curr * 10 <= n) {
        curr *= 10;
      } else {
        while (curr % 10 == 9 || curr == n)
          curr /= 10;
        ++curr;
      }
    }

    return ans;
  }
};

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class Solution {
  public List<Integer> lexicalOrder(int n) {
    List<Integer> ans = new ArrayList<>();
    int curr = 1;

    while (ans.size() < n) {
      ans.add(curr);
      if (curr * 10 <= n) {
        curr *= 10;
      } else {
        while (curr % 10 == 9 || curr == n)
          curr /= 10;
        ++curr;
      }
    }

    return ans;
  }
}

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class Solution:
  def lexicalOrder(self, n: int) -> list[int]:
    ans = []
    curr = 1

    while len(ans) < n:
      ans.append(curr)
      if curr * 10 <= n:
        curr *= 10
      else:
        while curr % 10 == 9 or curr == n:
          curr //= 10
        curr += 1

    return ans