Python 中两个链表的交集

pythonserver side programmingprogramming更新于 2025/10/31 21:32:17

假设我们有两个链表 A 和 B，这两个链表中的元素很少。我们需要返回交点的引用。输入为intersectionVal = 8，A = [4,1,8,4,5]，B = [5,0,1,8,4,5]，skipA = 2和skipB = 3，它们用于从A中跳过2个元素并从B中跳过3个元素。

为了解决这个问题，我们将遵循以下步骤−

定义一个名为d的映射
当headA不为空时
- d[headA] := 1
- headA := next of headA
当headB不为空时
- 如果headB在d中
  - 返回headB
- headB := next of headB
返回 null

示例

让我们看看下面的实现以便更好地理解 −

class ListNode:
   def __init__(self, data, next = None):
      self.data = data
      self.next = next
class Solution(object):
   def getIntersectionNode(self, headA, headB):
      """
      :type head1, head1: ListNode
      :rtype: ListNode
      """
      dict = {}
      while headA:
         dict[headA]=1
         headA = headA.next
      while headB:
         if headB in dict:
            return headB
         headB = headB.next
      return None
headA = ListNode(4)
headB = ListNode(5)
Intersect = ListNode(8, ListNode(4, ListNode(5)))
headA.next = ListNode(1, Intersect)
headB.next = ListNode(0, ListNode(1, Intersect))
ob1 = Solution()
op = ob1.getIntersectionNode(headA, headB)
print("Intersection:",op.data)

输入

headA = ListNode(4)
headB = ListNode(5)
Intersect = ListNode(8, ListNode(4, ListNode(5)))
headA.next = ListNode(1, Intersect)
headB.next = ListNode(0, ListNode(1, Intersect))