LeetCode 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.

时间复杂度是O(n)的话，刚开始看，一下子没了思路。于是参考了陆草纯的思路，如下：

可以将A,B两个链表看做两部分，交叉前与交叉后。

交叉后的长度是一样的，因此交叉前的长度差即为总长度差。

只要去除这些长度差，距离交叉点就等距了。

为了节省计算，在计算链表长度的时候，顺便比较一下两个链表的尾节点是否一样，

若不一样，则不可能相交，直接可以返回NULL

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {if (headA == null || headB == null) return null;ListNode tmpA = headA;ListNode tmpB = headB;int lenA = 0;int lenB = 0;while (tmpA != null){tmpA = tmpA.next;lenA++;}while (tmpB != null){tmpB = tmpB.next;lenB++;}if (tmpA != tmpB) return null;int dif = lenA - lenB;tmpA = headA;tmpB = headB;while (dif > 0){tmpA = tmpA.next;dif--;}while (dif < 0){tmpB = tmpB.next;dif++;}while (tmpA != tmpB){tmpA = tmpA.next;tmpB = tmpB.next;}return tmpA;}