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.

可以將A,B兩個鏈表看做兩部分，交叉前與交叉後。例子中

交叉前A： a1 → a2

交叉前B: b1 → b2 → b3

交叉後AB一樣： c1 → c2 → c3

所以 ，交叉後的長度是一樣，所以，交叉前的長度差即為總長度差。

只要去除長度差，距離交叉點就等距了。

 1    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 2         if(headA == null || headB == null) return null;
 3         ListNode tempA = headA;//
 
 計算鏈表長度用
 4          ListNode tempB = headB;
 5          int len_A = 1;
 6          int len_B = 1;
 7          while (tempA.next != null) {tempA = tempA.next; len_A++;}// 計算鏈表長度
 8          while (tempB.next != null) {tempB = tempB.next; len_B++;}
 9          int diff ;//長度差
10          //去除長度差
11          if(len_A > len_B){     
12              diff = len_A - len_B;
13              while(diff > 0){
14                  headA = headA.next;
15                    diff--;
16              }
17          }
18          else if (len_A < len_B){
19              diff = len_B - len_A;
20              while(diff > 0){
21                  headB = headB.next;
22                    diff--;
23              }
24          }
25         while(headA != headB){
26             headA = headA.next;
27             headB = headB.next;
28         }
29         return headA;
30     }

參考：

http://www.cnblogs.com/ganganloveu/p/4128905.html

[LeetCode]160.Intersection of Two Linked Lists