Leetcode 160 Intersection of Two Linked Lists(快慢指针)

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.

题目解析：

Leetcode 里面好多链表的题都可以用快慢指针来解.

这道题的思路就是先遍历两个lists然后得到两个lists的长度, 然后让长的那个lists先走Math.abs(lengthA-lengthB)步, 然后一起走判断下一个节点是不是相等.

注意其中有一步就是两个lists遍历之后要判断最后一个节点是不是相等, 不等就可以直接返回null啦~

 1 public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 2     if(headA == null && headB == null)
 3         return null;
 4     ListNode node1 = headA;
 5     ListNode node2 = headB;
 6
 7     int lengthA = 1;
 8     int lengthB = 1;
 9     while(node1 != null){
10         lengthA++;
11         node1 = node1.next;
12     }
13     while(node2 != null){
14         lengthB++;
15         node2 = node2.next;
16     }
17
18     if(node1 != node2)    //先判断一下
19         return null;
20     else{
21         int count = Math.abs(lengthA - lengthB);
22         if(lengthA > lengthB){    //此处处理要注意
23             node1 = headA;
24             node2 = headB;
25         }else{
26             node2 = headA;
27             node1 = headB;
28         }
29         for(int i = 0; i < count; i++){
30             node1 = node1.next;
31         }
32         while(node1 != null && node2 != null & node1 != node2){
33             node1 = node1.next;
34             node2 = node2.next;
35         }
36
37     }
38     return node1;
39 }