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authorJack Sangdahl <[email protected]>2025-03-10 00:36:13 -0600
committerJack Sangdahl <[email protected]>2025-03-10 00:36:13 -0600
commit7c60dee83b436e8e8a667bf9f191beced8de4fc3 (patch)
treeeb7a24ee2067f92245c57bb118e85a1baf15df50 /other/lexicographical_numbers.cpp
parent5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff)
downloadleetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.tar.gz
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chore: move cpp and rust code to directory
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+// 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");
+}