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