Sequence element cross-specific algorithm

Concept

In Python, elements in data types such as list,tuple can be duplicated, unlike set types (there are no duplicate elements in set). In this, there must be a corresponding algorithm to verify whether the sequence has duplicate elements.

Achieve a

def  unique1 (S):  for  J 
     
      in 
      range  (len  (S)): for  K in  range  (J +  1 ,                span class= "bu" >len  (S)): if  s[j] ==  S[k]: return     false     # found duplicate pair  return                  true  # if we reach this, elements were unique   

The idea of implementing one is simple, and the outer loop iterates through each value in the sequence, from left to right. S[0] and s[1] ... S[N-1] [comparison], s[1] and s[2] ... S[n-1] is compared, until s[n-2] and s[n-1] are compared, then the traversal ends.
The number of comparisons mentioned above is (n-1) + (n-2) + ... + 2 + 1. Therefore, the total time complexity is O ($ n^2 $).

Implementation two

def  unique2 (S):  temp =  
     
      sorted 
      (S) for  J in  range  (1 , len  (temp)): if  temp[j - 1 ] ==  temp[j]: return   span class= "va" >false  # found duplicate pair  return  true  # if we reach this, elements were unique   

As the code example shows, the first sort is sorted by Python's built-in sort function, by default, from small to large. If the same element exists in the sequence before sorting, the position of the same element is adjacent after sorting. For example, a = [5, 3, 2, 3, 4],b = sorted (a), then B = [2, 3, 3, 4, 5]. In this case, when iterating through a sequence, you only need to compare adjacent elements for equality. If you compare from left to right, there are no equal elements, the elements in the sequence are different.

• Built-in function sorted, which produces a copy of the original sequence after ordering. Its maximum time complexity is O (n$ \log $n). The time complexity of the sorting algorithm is analyzed separately.
• In a looping statement, the traversal sequence is n, and its time complexity is O (n). For this, the time complexity of the entire algorithm is O (n$ \log $n).

Sequence element cross-specific algorithm