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交叉链表求交点
实现
def getIntersectionNode(self, headA, headB):
"""
:type head1, head1: ListNode
:rtype: ListNode
"""
if headA is not None and headB is not None:
pa, pb = headA, headB
while pa is not pb:
pa = pa.next if pa is not None else headB
pb = pb.next if pb is not None else headA
return pa
关于代码的一点解释
pa、pb 2 个指针的初始位置是 headA、headB 头结点,pa、pb 两个指针一直往后遍历。
直到 pa 指针走到链表的末尾,然后 pa 指向 headB;
直到 pb 指针走到链表的末尾,然后 pb 指向 headA;
然后 pa、pb 再继续遍历;
每次 pa、pb 指向 None,则将 pa、pb 分别指向 headB、headA。
循环的次数越多,pa、pb 的距离越接近,直到 pa 等于 pb。
参考文献
- 160. Intersection of Two Linked Lists - LeetCode;
- https://leetcode.com/problems/intersection-of-two-linked-lists/discuss/49846/Python-solution-for-intersection-of-two-singly-linked-lists;
- https://leetcode.com/problems/intersection-of-two-linked-lists/discuss/49798/Concise-python-code-with-comments;
- https://leetcode.com/submissions/detail/194298767/。