第20669题
采用分治思想实现的最大连续子段和代码的时间复杂度为多少?
int solve(vector<int>& a, int l, int r){ if(l == r) return a[l];

  int mid = l + (r - l) / 2;

  int left = solve(a, l, mid);
  int right = solve(a, mid + 1, r);

  int sum = 0, lmax = INT_MIN;
  for(int i = mid; i >= l; i--){
   sum += a[i];
   lmax = max(lmax, sum);
  }

  sum = 0;
  int rmax = INT_MIN;
  for(int i = mid + 1; i <= r; i++){
   sum += a[i];
   rmax = max(rmax, sum);
  }

  return max({left, right, lmax + rmax});
 }
A

O(n²)

B

O(nlogn)

C

O(logn)

D

O(n)