Concise O(N) C++ Solution


  • 0
    D
        int thirdMax(vector<int>& nums) {
            vector<long long> vec = { LONG_MIN, LONG_MIN, LONG_MIN };
            for (int i = 0; i < nums.size(); i++) {
                if (nums[i]>vec[0] && nums[i]!=vec[1] && nums[i]!=vec[2]) {
                    vec[0] = nums[i];
                    sort(vec.begin(), vec.end());
                }
            }
            return vec[0] == LONG_MIN ? vec[2] : vec[0];
        }
    

    We have a sort inside the loop, but time complexity is still O(N) as we're sorting only 3 numbers no matter the size of the input.


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