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.

没想到什么特别的，上hash。。空间O(N),不完美

public ListNode getIntersectionNode(ListNode headA, ListNode headB){ListNode n=headA;HashSet<ListNode> hashset=new HashSet<>(512);while(n!=null){hashset.add(n);n=n.next;}n=headB;while(n!=null){if(hashset.contains(n))return n;n=n.next;}return null;}

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空间O(1)的，摘自：https://leetcode.com/articles/intersection-two-linked-lists/

Approach #3 (Two Pointers) [Accepted]

• Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
• When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
• If at any point pA meets pB, then pA/pB is the intersection node.
• To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
• If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.

Complexity Analysis

• Time complexity : $O(m+n)$.
• Space complexity : $O(1)$.

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {    int lenA = length(headA), lenB = length(headB);    // move headA and headB to the same start point    while (lenA > lenB) {        headA = headA.next;        lenA--;    }    while (lenA < lenB) {        headB = headB.next;        lenB--;    }    // find the intersection until end    while (headA != headB) {        headA = headA.next;        headB = headB.next;    }    return headA;}private int length(ListNode node) {    int length = 0;    while (node != null) {        node = node.next;        length++;    }    return length;}