*Intersection of Two Linked Lists

题目:

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.

思路 方法超屌!!!

1. 得到2个链条的长度。

2. 将长的链条向前移动差值(len1 - len2)

3. 两个指针一起前进,遇到相同的即是交点,如果没找到,返回null.

相当直观的解法。空间复杂度O(1), 时间复杂度O(m+n)

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
14         if (headA == null || headB == null) {
15             return null;
16         }
17         
18         ListNode cur = headA;
19         int len1 = getLen(headA);
20         int len2 = getLen(headB);
21         
22         int cnt = Math.abs(len1 - len2);
23         
24         // cut the longer list.
25         if (len1 > len2) {
26             while (cnt > 0) {
27                 headA = headA.next;
28                 cnt--;
29             }
30         } else {
31             while (cnt > 0) {
32                 headB = headB.next;
33                 cnt--;
34             }
35         }
36             
37         while (headA != null) {
38             if (headA == headB) {
39                 return headA;
40             }
41             
42             headA = headA.next;
43             headB = headB.next;
44         }
45         
46         return null;
47     }
48     
49     public int getLen(ListNode head) {
50         int cnt = 0;
51         while (head != null) {
52             head = head.next;
53             cnt++;
54         }
55         
56         return cnt;
57     }
58 }

reference: http://www.cnblogs.com/yuzhangcmu/p/4128794.html

原文地址:https://www.cnblogs.com/hygeia/p/4759571.html