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;    }};