160. 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.

思路：

找链表交界，很类似Linked List Cycle II那题，方法也是类似的双指针相遇法。分两步走：

1. 如何判断两链表是否相交？

两链表相交则他们必然有共同的尾节点。所以遍历两个链表，找到各自的尾节点，如果tailA!=tailB则一定不相交，反之则相交。

2. 如何判断两链表相交的起始节点？

在第1步判断相交时可以顺带计算两链表的长度lenA和lenB。让长的链表的head先走abs(lenA-lenB)步，然后和短链表的head一起走，直到两者相遇，即为要找的节点。

class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(!headA || !headB) return NULL;
int lenA = 0, lenB = 0;
ListNode *tailA = headA, *tailB = headB;

while(tailA->next) {
tailA = tailA->next;
lenA++;
}
while(tailB->next) {
tailB = tailB->next;
lenB++;
}
if(tailA!=tailB) return NULL;

if(lenA>lenB) {
for(int i=0; i<lenA-lenB; i++)
headA = headA->next;
}
else {
for(int i=0; i<lenB-lenA; i++)
headB = headB->next;
}

while(headA!=headB) {
headA = headA->next;
headB = headB->next;
}

return headA;
}
};