题目：题目链接（简单）
标签：链表、链表-相交链表、链表-双指针、链表-快慢针
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N ) O(N) O(N) | O ( N ) O(N) O(N) | 216ms (25.96%) |
| Ans 2 (Python) | O ( N ) O(N) O(N) | O ( 1 ) O(1) O(1) | 184ms (75.53%) |
| Ans 2 (Python) | O ( N ) O(N) O(N) | O ( 1 ) O(1) O(1) | 180ms (84.94%) |
解法一（使用Python的list直接实现）：
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
hashmap = set()
while headA:
hashmap.add(headA)
headA = headA.next
while headB:
if headB in hashmap:
return headB
headB = headB.next
解法二（双指针；因为双指针经过长度相同，因此一定会同时到达终点）：
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
if headA and headB:
nodeA = headA
nodeB = headB
while nodeA != nodeB:
nodeA = nodeA.next if nodeA else headB
nodeB = nodeB.next if nodeB else headA
return nodeA
解法三（双指针）：
def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
nodeA, nodeB = headA, headB
while nodeA != nodeB:
nodeA = nodeA.next if nodeA else headB
nodeB = nodeB.next if nodeB else headA
return nodeA
