Problem: 160. 相交链表

思路 & 解题方法

可以用hash做，也可以做一个技巧。

复杂度

时间复杂度:

添加时间复杂度, 示例： $O(n)$

空间复杂度:

添加空间复杂度, 示例： $O(n)$

哈希

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:hashTable = collections.defaultdict(int)while headA:hashTable[headA] += 1headA = headA.nextwhile headB:if headB in hashTable:return headBheadB = headB.nextreturn None

技巧

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Noneclass Solution:def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:A, B = headA, headBwhile A != B:A = A.next if A else headBB = B.next if B else headAreturn A