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author | Jack Sangdahl <[email protected]> | 2025-06-27 12:23:02 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2025-06-27 12:23:02 -0600 |
commit | a96b65cbfad7f62f3db6b2eab105e4e25c0414c5 (patch) | |
tree | e2a2b3f49f9fb741be81415266ac93e74cff42ec /other/single_number2.cpp | |
parent | 694ad5ed71edcb1deccab54b72d706b779918276 (diff) | |
download | leetcode-a96b65cbfad7f62f3db6b2eab105e4e25c0414c5.tar.gz leetcode-a96b65cbfad7f62f3db6b2eab105e4e25c0414c5.zip |
chore: remove old cpp and rust solutions
Diffstat (limited to 'other/single_number2.cpp')
-rw-r--r-- | other/single_number2.cpp | 29 |
1 files changed, 0 insertions, 29 deletions
diff --git a/other/single_number2.cpp b/other/single_number2.cpp deleted file mode 100644 index 907fee0..0000000 --- a/other/single_number2.cpp +++ /dev/null @@ -1,29 +0,0 @@ -// 137. Single Number II -#include "leetcode.h" - -/* - * given an integer array 'nums' where every element appears three times except - * for one, which appears exactly once. find the single element and return it. - * you must implement a solution with a linear runtime complexity and use only - * constant extra space. - */ - -class Solution { -public: - int singleNumber(vector<int> &nums) { - unordered_map<int, int> um; - for (auto i : nums) - um[i]++; - for (auto i : um) - if (i.second == 1) - return i.first; - return -1; - } -}; - -int main() { - Solution obj; - vector<int> n1 = {2, 2, 3, 2}, n2 = {0, 1, 0, 1, 0, 1, 99}; - printf("%d\n", obj.singleNumber(n1)); // expect: 3 - printf("%d\n", obj.singleNumber(n2)); // expect: 99 -} |