基于Python的编码带平衡训练

Question

A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape. Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1]. The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])| In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

def solution(A):
N = len(A)
my_list = []
for i in range(1, N):
    first_tape = sum(A[:i - 1]) + A[i]
    second_tape = sum(A[i - 1:]) + A[i]
    difference = abs(first_tape - second_tape)
    my_list.append(difference)
print(min(my_list))
return min(my_list)

我的解决方案的正确率为100%，但性能为0%。 我想应该是O（N），但我的时间复杂度是O（N*N）。 谁能给我建议一下吗？在

Answer 1

您可以将您的代码更改为类似下面这样的代码，以使其具有复杂性O(N)。在

def solution(A):          
    s = sum(A)
    m = float('inf')
    left_sum = 0
    for i in A[:-1]:
        left_sum += i
        m = min(abs(s - 2*left_sum), m)
    return m

Answer 2

我的java代码 O（N）

class Solution {
public int solution(int[] arr) {

   int sum = 0;

   for(int i = 0; i<arr.length; i++){
       sum = sum + arr[i];
   }



  int  minSum = 100000;
  int  tempSum = 0;
  int previousSum = 0;
   for(int i = 0; i<arr.length-1; i++){

       previousSum = previousSum + arr[i];

      tempSum = Math.abs(previousSum - (sum - previousSum));

      if(minSum > tempSum){
          minSum = tempSum;
      }


   }

   return minSum;
}

}

Answer 3

为了回答您的问题，它是O（n*n），因为sum()函数是O（n）时间复杂度，并且您在一个for循环中调用它，这个循环也是O（n）。在

因此算法的时间复杂度为O（N*N）