给定两个 lists/vectors，其中一个是另一个的子集，我怎样才能以最有效的方式对齐它们？

Question

给定两个 lists/vectors，其中一个是另一个的子集，我怎样才能以最有效的方式对齐它们？对于上下文，我正在对矩阵进行切片（在 C++ 中），我还需要对标签进行切片。由于这些列表将是标签，因此它们在列表中应被视为唯一。

import unittest
from typing import List, Tuple 

def doAlignment(list1: List[int], list2: List[int]) -> Tuple[int, int]:
    pass

#The elements in the lists are labels, so they should be unique - i.e. we won't 
# ever have [5, 5, 6, 7, 8].

class AlignTests(unittest.TestCase):
    
    def test1(self):
        v1 = [1, 2, 3, 4, 5, 6]
        v2 = [1, 2]
        first, last = doAlignment(v1, v2)
        self.assertEqual((0, 1), (first, last))
        
    def test2(self):
        v1 = [1, 2, 3, 4, 5, 6]
        v2 = [2, 3, 4]
        first, last = doAlignment(v1, v2)
        self.assertEqual((1, 3), (first, last))

    def test3(self):
        v1 = [7, 3, 4, 6, 9]
        v2 = [3, 4, 6]
        first, last = doAlignment(v1, v2)
        self.assertEqual((1, 3), (first, last))
        
    def test4(self):
        v1 = [7, 3, 4, 6, 9]
        v2 = [3, 4, 6, 9]
        first, last = doAlignment(v1, v2)
        self.assertEqual((1, 4), (first, last))

Answer 1

由于list1中的值是唯一的，只需获取list2的第一项的索引即可获得第一个元组值，为第二个添加len(list2)-1：

def doAlignment(list1: List[int], list2: List[int]) -> Tuple[int, int]:
    a = list1.index(list2[0])
    return (a, a+len(list2)-1)

示例：

>>> doAlignment([7, 3, 4, 6, 9], [3, 4, 6, 9])
(1, 4)

给定两个 lists/vectors，其中一个是另一个的子集，我怎样才能以最有效的方式对齐它们？

Given two lists/vectors, one of which is a subset of another, how can I align them in the most efficient way possible?

python

list

algorithm

vector

alignment