code1 [5,19,8,1]输出3 19/2/2+8/2共计3次 遍历列表每次对最大值除2,三个测试案例都能过复杂度应该也还可以 def solution(A): A.sort() target = sum(A)/2 A = A[::-1] if len(A)<=2: return len(A) B = A[:] res,count = 0,0 while res<target: temp = [] m = max(B)/2 res += m ...