Intersection of Two Linked Lists

160. Intersection of Two Linked Lists

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Question

Total Accepted: 65866 Total Submissions: 218868 Difficulty: Easy

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.

双指针法  pa， pb  

遍历链表A，记录其长度lengthA，遍历链表B，记录其长度lengthB。

因为两个链表的长度可能不相同，比如题目所给的case，lengthA=5，lengthB=6，则作差得到 lengthB- lengthA=1，将指针pb从链表B的首节点开始走1步，即指向了第二个节点，pa指向链表A首节点，然后它们同时走，每次都走一步，当它们相等时，就是交集的节点。

时间复杂度O(lengthA+lengthB)，空间复杂度O(1)。双指针法的代码如下：

/** * Created by jason on 2016/3/11. */class Solution75 {    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        if (headA == null || headB == null) return null;        ListNode pa = headA;        ListNode pb = headB;        int lengthA=0, lengthB=0;        while (pa != null) {            pa = pa.next;            lengthA++;        }        while (pb != null) {            pb = pb.next;            lengthB++;        }        if (lengthA <= lengthB) {            int n = lengthB - lengthA;            pa = headA; pb = headB;            while (n != 0) {                pb = pb.next;                n--;            }        }else {            int n = lengthA - lengthB;            pa = headA; pb = headB;            while (n!=0) {                pa = pa.next;                n--;            }        }        while (pa != pb) {            pa = pa.next;            pb = pb.next;        }        return pa;    }}public class LC75 {}