LeetCode#160 Intersection of Two Linked Lists

Problem Definition:

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.

Solution 1: Seems a little bit long, but it's straightforward.

 1 def getIntersectionNode( headA, headB):
 2         l1,l2=0,0
 3         ha,hb=headA,headB
 4         while ha!=None:
 5             ha=ha.next
 6             l1+=1
 7         while hb!=None:
 8             hb=hb.next
 9             l2+=1
10         ha,hb=headA,headB   #throwback
11         if l1>l2:
12             sub=l1-l2
13             while sub>0:
14                 ha=ha.next
15                 sub-=1
16         else:
17             sub=l2-l1
18             while sub>0:
19                 hb=hb.next
20                 sub-=1
21         while ha!=hb:
22             ha=ha.next
23             hb=hb.next
24         return ha

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

 Solution 2：利用同样的原理，不同的解释。

1）同时遍历两个链表。ha遍历headA所指代链表,hb遍历headB所指代链表。

2）当ha到达链表尾时，将其指向headB，继续遍历；当hb到达链表尾时，将其指向headA，继续遍历。

3）当ha==hb时，返回ha。

　　看下面的栗子：

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

ha从a1开始走到c3，然后指向b1。此时hb正指向c3。
下一时刻，ha到达b2，而hb到达a1。如此便达到了类似前一种解法中的（使得两链表从到末节点距离相同的两个节点开始遍历）效果
继续遍历就会到达c1。

在以上过程中，每一步（ha和hb都移动后）都判断ha==hb，一旦成立则立即返回。
因此会有以下不同情况
1）两表有公共子链，但两表不等长，则在第二轮遍历时会找到交叉节点
2）两表有公共子链，且两表等长，则在第一轮遍历时就能找到交叉节点
3）两表没有公共子链，则在第一轮或者第二轮遍历中同时变为空，相等，返回空

代码：

1 def getIntersectionNode( headA, headB):
2     ha,hb=headA,headB
3     while ha!=hb:
4         ha=ha.next if ha!=None else headB
5         hb=hb.next if hb!=None else headA
6     return ha