LeetCode.160.Intersection of Two Linked Lists 两链表的交点

题目描述

160 Intersection of Two Linked Lists
https://leetcode-cn.com/problems/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:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

 
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.

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解题过程

我的想法:
同时开始遍历两个链表，其中一个结束后开始计数，得到两个链表长度差m，此时也知道了哪个链表更长。
较长的链表先走m步，然后较短的链表开始，不断比较结点是否相等，遇到的第一个相等结点就是交点。

看了题解才知道 时间 O(n) 空间 O(1) 的解法，很巧妙，自己想不出来。
双指针法，分别从两个链表头部开始，消除长度差：

法一：每个指针走到结尾后跳到另一个链表头部继续，若有交点则第二次遍历肯定会相遇，若走完都没相遇，肯定无交点。 法二：第一个指针到达末端时，另一个与末端距离是两链表的长度差，记为d，此时也知道了哪个链表长。然后，长链表指针先走d步消除长度差，若有交点肯定相遇，否则不相遇。

private static class SolutionV2020 {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (null == headA || null == headB) {
            return null;
        }
        ListNode pA = headA, pB = headB;

        // 双指针法，每个指针走到结尾后跳到另一个链表头部继续，消除长度差，若有交点则第二次遍历肯定会相遇
        // 若走完都没相遇，肯定无交点
        while (null != pA || null != pB) {
            if (pA == pB) {
                return pA;
            }
            pA = null != pA ? pA.next : headB;
            pB = null != pB ? pB.next : headA;
        }
        return null;
    }
}

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GitHub代码

algorithms/leetcode/leetcode/_160_IntersectionOfTwoLinkedLists.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_160_IntersectionOfTwoLinkedLists.java

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