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.
- Difficulty: Easy
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.
要求:这个题要返回两个链表交集的第一个结点。时间复杂度为O(n) ,空间复杂度为O(1)
下面提供两种比较经典的思路:
方案一:1、 将两个链表遍历一次,得到每个链表的长度,和每个链表的最后一个结点,如果两个链表的最后一个结点不同,直接return null
2、根据两个链表的差值,将长的链表的指针先移动的差值位置,在同时向下遍历,知道找到第一个相同的结点,返回。
由于方案一较为常规,这里就不提供具体实现代码了。下面介绍一种特殊的解决方案。
方案二:
具体实现方式是从两个链表的头结点同时开始遍历,当一个链表的next为空时,从另一个链表的头开始遍历。也就是说两个链表遍历的总长度是相同的。
其中必然会出现交点。第一个相同的结点即为所求。
已上面两链表为例:遍历过程如下:
A: a1 -> a2 -> c1 -> c2 -> c3 -> b1 -> b2 -> b3 ->c1 -> c2 -> c3
B: b1 -> b2 -> b3 -> c1 -> c2 -> c3 -> a1 -> a2 ->c1 -> c2 -> c3
故c1为第一个交集的第一个结点。
public ListNode getIntersectionNode(ListNode headA, ListNode headB) { //boundary check if(headA == null || headB == null) return null; ListNode a = headA; ListNode b = headB; //if a & b have different len, then we will stop the loop after second iteration while( a != b){ //for the end of first iteration, we just reset the pointer to the head of another linkedlist a = a == null? headB : a.next; b = b == null? headA : b.next; } return a;}
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