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

For example, the following two linked lists:

begin to intersect at node c1.

 

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3Output: Reference of the node with value = 8Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

 

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1Output: Reference of the node with value = 2Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

 

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2Output: nullInput Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.Explanation: The two lists do not intersect, so return null.

 

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.

 

方法一：双指针

分析：链表a的长度a+c，链表b的长度：b+c 。链表重合部分的长度是：c

那么如果指向a的指针走过长度是：a+c+b

　　　　指向b的指针走过的长度是：b+c+a

的时候两个指针指向 的下一个结点就是第一个公共链的结点。（不然就是空节点，没有公共链）

 

时间复杂度：o（n+m）                               空间复杂度：o(1)

/*public class ListNode {    int val;    ListNode next = null;    ListNode(int val) {        this.val = val;    }}*/public class Solution {    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {        ListNode headA=pHead1;        ListNode headB=pHead2;        while (pHead1!=pHead2){            pHead1=(pHead1==null)?headB:pHead1.next;            pHead2=(pHead2==null)?headA:pHead2.next;        }        return pHead1;    }}

 方法二：利用hashset特性

hashset用于存储对象，用它存储pHead1链表所有结点。虽然它是无序的，不保证存入和取出的顺序相同，但是遍历pHead2的时候是有序的。遍历到第一个结点时就会立刻返回。

import java.util.HashSet;/*public class ListNode {    int val;    ListNode next = null;    ListNode(int val) {        this.val = val;    }}*/public class Solution {    public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {        HashSet
     
       set=new HashSet
      
       ();        while (pHead1!=null){            set.add(pHead1);            pHead1=pHead1.next;        }        while (pHead2!=null){            if(set.contains(pHead2))                return pHead2;            pHead2=pHead2.next;        }        return pHead2;    }}