函数f()最坏情况下的时间复杂度为？

 #include <bits/stdc++.h>
 using namespace std;
 vector<int> nums1, nums2;
 int m, n, t;
 int f(int k) {
     int index1 = 0, index2 = 0;
     while (true) {
         if (index1 == m) {
             return nums2[index2 + k - 1];
         }
         if (index2 == n) {
             return nums1[index1 + k - 1];
         }
         if (k == 1) {
             return min(nums1[index1], nums2[index2]);
         }
         int newIndex1 = min(index1 + k / 2 - 1, m - 1);
         int newIndex2 = min(index2 + k / 2 - 1, n - 1);
         int pivot1 = nums1[newIndex1];
         int pivot2 = nums2[newIndex2];
         if (pivot1 <= pivot2) {
             k -= newIndex1 - index1 + 1;
             index1 = newIndex1 + 1;
         } else {
             k -= newIndex2 - index2 + 1;
             index2 = newIndex2 + 1;
         }
     }
 }
 
 int main(){
     cin >> m >> n;
     for (int i = 0; i <= m-1; i++){
         cin >> t;
         nums1.push_back(t);
     }
     for (int i = 0; i <= n-1; i++){
         cin >> t;
         nums2.push_back(t);
     }
     int totalLength = nums1.size() + nums2.size();
     if (totalLength & 1) {
         cout << f((totalLength + 1) / 2);
     } else {
         cout << (f(totalLength / 2) + f(totalLength / 2 + 1)) / 2.0;
     }
     return 0;
 }