【Leetcode】160. Intersection of Two Linked Lists（链表结合点）

骑猪看日落 2022-02-03 05:05 141阅读 0赞

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

[![160_statement.png][]][160_statement.png]

begin to intersect at node c1.

**Example 1:**

[![160_example_1.png][]][160_example_1.png]

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:**

[![160_example_2.png][]][160_example_2.png]

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:**

[![160_example_3.png][]][160_example_3.png]

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.

题目大意：

求解两个链表的结合点，保证算法复杂度在O(n)，不开辟新内存。其中包含无连接点两个链表。

解题思路：

一张图解决

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzI5NjAwMTM3_size_16_color_FFFFFF_t_70][]

如果有连接点则从两个链表的头部出发，连接另一个的头部，经过相同的步骤即可同时到达交叉点。

/**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
            ListNode *ha=headA, *hb=headB;
            while(ha && hb){
                if(ha == hb){
                    return ha;
                }
                ha = ha->next;
                hb = hb->next;
                if(!ha && !hb)  return NULL;
                if(!ha) ha = headB;
                if(!hb) hb = headA;
            }
     
            return NULL;
        }
    };

[160_statement.png]: https://assets.leetcode.com/uploads/2018/12/13/160_statement.png
[160_example_1.png]: https://assets.leetcode.com/uploads/2018/12/13/160_example_1.png
[160_example_2.png]: https://assets.leetcode.com/uploads/2018/12/13/160_example_2.png
[160_example_3.png]: https://assets.leetcode.com/uploads/2018/12/13/160_example_3.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzI5NjAwMTM3_size_16_color_FFFFFF_t_70]: /images/20220203/6cd40376d9ff412484a7777ae709a14b.png