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author | Jack Sangdahl <jack@pngu.org> | 2025-03-10 00:36:13 -0600 |
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committer | Jack Sangdahl <jack@pngu.org> | 2025-03-10 00:36:13 -0600 |
commit | 7c60dee83b436e8e8a667bf9f191beced8de4fc3 (patch) | |
tree | eb7a24ee2067f92245c57bb118e85a1baf15df50 /lexicographical_numbers.cpp | |
parent | 5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff) | |
download | leetcode-master.tar.gz leetcode-master.zip |
Diffstat (limited to 'lexicographical_numbers.cpp')
-rw-r--r-- | lexicographical_numbers.cpp | 43 |
1 files changed, 0 insertions, 43 deletions
diff --git a/lexicographical_numbers.cpp b/lexicographical_numbers.cpp deleted file mode 100644 index ab48ce5..0000000 --- a/lexicographical_numbers.cpp +++ /dev/null @@ -1,43 +0,0 @@ -// 386. Lexicographical Numbers -#include "leetcode.h" - -/* - * given an integer 'n', return all the numbers in the range '[1, n]' sorted in - * lexicographical order. you must write an algorithm that runs in O(n) time - * complexity and uses O(1) extra space. - */ - -class Solution { -public: - vector<int> ans; - vector<int> lexicalOrder(int n) { - ans.resize(n); - int pos = 0; - for (int i = 1; i <= 9 && i <= n; i++) - helper(i, n, pos); - return ans; - } - -private: - void helper(int i, int n, int pos) { - ans[pos++] = i; - int tmp; - for (int j = 0; j <= 9; j++) { - tmp = i * 10 + j; - if (tmp > n) - break; - helper(tmp, n, pos); - } - } -}; - -int main() { - Solution obj; - vector<int> lo1 = obj.lexicalOrder(13), lo2 = obj.lexicalOrder(2); - for (int i = 0; i < lo1.size(); i++) - printf("%d ", lo1[i]); // expect: [1,10,11,12,13,2,3,4,5,6,7,8,9] - printf("\n"); - for (int i = 0; i < lo2.size(); i++) - printf("%d ", lo2[i]); // expect: [1,2] - printf("\n"); -} |