Intersection of Two Linked Lists

时间:2024-04-06 14:03:29

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.

Credits:

Special thanks to @stellari for adding this problem and creating all test cases.

算法:

1)把其中的一条链首尾相连,组成一个环,这样问题就转化成:判断一条链表是否有环,如果有,求环的入口问题。

2)把两个链表都首尾颠倒顺序,然后比较。比较完之后,再颠倒把链表的顺序,恢复原样。

算法1的代码:

c++

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(!headA||!headB)
return NULL;
ListNode *p = headB;
while(p->next){
p=p->next;
}
ListNode *tailB = p;
tailB->next=headB; ListNode *p1,*p2;
p1=headA;
p2=headA;
while(p1){
p1=p1->next;
p2=p2->next;
if(p1){
p1=p1->next;
}else{
tailB->next = NULL;
return NULL;
}
if(p1==p2){
break;
}
}
if(p1==NULL){
tailB->next = NULL;
return NULL;
}
p2=headA;
while(p1!=p2){
p2=p2->next;
p1=p1->next;
}
tailB->next = NULL;
return p2; }
};