Leetcode: 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.
将一个链表的表尾指向另一个链表的表头，问题简化为求链表是否有环，如果有环求链表的第一个环节点。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL) {
return NULL;
}

ListNode *tailA = headA;
while (tailA->next != NULL) {
tailA = tailA->next;
}
tailA->next = headB;

ListNode *fast = headA;
ListNode *slow = headA;
while (fast != NULL && fast->next != NULL) {
fast = fast->next->next;
slow = slow->next;
if (slow == fast) {
break;
}
}
if (slow != fast) {
tailA->next = NULL;
return NULL;
}

ListNode *curA = headA;
while (curA != slow) {
curA = curA->next;
slow = slow->next;
}
tailA->next = NULL;

return slow;
}
};

另一种比较巧妙的思想，按照同样的节奏遍历两个链表，并记录尾节点；当某个链表到尾节点时继续从另一个链表的头结点遍历。这样如果尾节点不同则不相交，如果相交则第一个相同的节点必定同时到达
- 遍历的节点数目相同。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL) {
return NULL;
}

ListNode *iterA = headA;
ListNode *iterB = headB;
ListNode *tailA = NULL;
ListNode *tailB = NULL;
while (iterA != NULL && iterB != NULL && iterA != iterB) {
if (iterA->next == NULL) {
tailA = iterA;
iterA = headB;
}
else {
iterA = iterA->next;
}
if (iterB->next == NULL) {
tailB = iterB;
iterB = headA;
}
else {
iterB = iterB->next;
}

if (tailA != NULL && tailB != NULL && tailA != tailB) {
return NULL;
}
}

return iterA;
}
};