160. Intersection of Two Linked Lists

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.

含义：两个列表有交集，求出他们的交集开始点

方法一：

 1     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 2         ListNode a = headA;
 3         ListNode b = headB;
 4         //if a & b have different len, then we will stop the loop after second iteration
 5         while( a != b){
 6             //for the end of first iteration, we just reset the pointer to the head of another linkedlist
 7             a = a == null? headB : a.next;
 8             b = b == null? headA : b.next;
 9         }
10         return a; //如果a不等于null，代表找到了交集的开始点  如果a等于null，说明两个列表没有交集

11 }

方法二：

 1 private int getListLength(ListNode head) {
 2         int length = 0;
 3         while (head != null) {
 4             length++;
 5             head = head.next;
 6         }
 7         return length;
 8 }
 9 
10 public class Solution {
11     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
12         int lengthA = getListLength(headA);
13         int lengthB = getListLength(headB);
14         while (lengthA > lengthB) {
15             lengthA--;
16             headA = headA.next;
17         }
18         while (lengthB > lengthA) {
19             lengthB--;
20             headB = headB.next;
21         }
22         while (headA != headB) {
23             headA = headA.next;
24             headB = headB.next;
25         }
26         return headA; //如果headA不等于null，代表找到了交集的开始点  如果headA等于null，说明两个列表没有交集

27 }

方法三：解决方法就是将A链表尾节点指向头结点形成一个环，检测B链表是否存在环，如果存在，则两个链表相交，而检测出来的依赖环入口即为相交的第一个点

 1     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 2         if (headA == headB || headA==null) return headA;
 3         ListNode temp = headA;
 4         while (temp.next!=null) temp = temp.next;
 5         temp.next = headA;
 6         ListNode slow = headB,fast = headB;
 7         while (fast!=null && fast.next !=null)
 8         {
 9             slow = slow.next;
10             fast = fast.next.next;
11             if (slow == fast)
12             {
13                 while (headB != slow)
14                 {
15                     headB = headB.next;
16                     slow = slow.next;
17                 }
18                 return slow;
19             }
20 
21         }
22         return null;
23     }