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-rw-r--r--lexicographical_numbers.cpp43
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");
-}