[Swift]LeetCode160. 相交链表 | Intersection of Two Linked Lists

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Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2↘c1 → c2 → c3↗
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.

编写一个程序，找到两个单链表相交的起始节点。

例如，下面的两个链表：

A:          a1 → a2↘c1 → c2 → c3↗
B:     b1 → b2 → b3

在节点 c1 开始相交。

注意：

如果两个链表没有交点，返回 nil.
在返回结果后，两个链表仍须保持原有的结构。
可假定整个链表结构中没有循环。
程序尽量满足 O(n) 时间复杂度，且仅用 O(1) 内存。

 1 class Solution {
 2     func getIntersectionNode(_ headA: ListNode?, _ headB: ListNode? ) -> ListNode? {
 3         if headA == nil || headB == nil {
 4             return nil
 5         }
 6
 7         var a = headA
 8         var b = headB
 9         while (a !== b) {
10             if a != nil {
11                 a = a?.next
12             } else {
13                 a = headB
14             }
15
16             if b != nil {
17                 b = b?.next
18             } else {
19                 b = headA
20             }
21         }
22         return a
23     }
24
25     public class ListNode: Equatable {
26
27         public var val: Int
28         public var next: ListNode?
29
30         public init(_ val: Int) {
31             self.val = val
32             self.next = nil
33         }
34
35         public static func ==(lhs: Solution.ListNode, rhs: Solution.ListNode) -> Bool {
36             return lhs.val == rhs.val && lhs.next == rhs.next
37         }
38     }
39 }

转载于:https://www.cnblogs.com/strengthen/p/10109542.html

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