2025/2/26 【双指针】LeetCode 面试题 02.07. 链表相交 【√】

面试题 02.07. 链表相交 - 力扣（LeetCode）

代码随想录

1. 解法1

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode):
        # 1. 分别求得两个链表的长度
        length_A = length_B = 0
        p_A = headA
        p_B = headB
        while p_A:
            length_A += 1
            p_A = p_A.next
        
        while p_B:
            length_B += 1
            p_B = p_B.next
        
        start_A = headA
        start_B = headB

        # 2.比较链表长度，找到使得两个链表等长的位置
        if length_A > length_B:
            gap = length_A - length_B
            while gap:
                start_A = start_A.next
                gap -= 1
        else:
            gap = length_B - length_A
            while gap:
                start_B = start_B.next
                gap -= 1
        
        while start_A != start_B:
            start_A = start_A.next
            start_B = start_B.next
        
        # 3.开始遍历
        return start_A
        

2.等比例法（来自代码随想录）

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode):
        # 0.处理边缘情况
        if not headA or not headB:
            return None
        
        # 1.分别在两个链表的头部初始化两个指针
        start_A = headA
        start_B = headB

        # 2.遍历两个链表直到指针相交
        while start_A != start_B:
            start_A = start_A.next if start_A else headB
            start_B = start_B.next if start_B else headA
        
        # 3.如果相交，指针将位于交点节点，如果没有交点，值为None
        return start_A
        

注意不要写成：

start_A = start_A.next if start_A.next else headB
start_B = start_B.next if start_B.next else headA

这样代码陷入死循环

感受：这道题是【简单题】，如果线性思维去解题，解法1和解法2真不容易想到。需要像做数学的找规律题一样去做。有了这个思路，就会很容易找到规律。