LeetCode 160： Intersection of Two Linked Lists

2024年2月18日 113次阅读

题目链接：

https://leetcode.com/problems/intersection-of-two-linked-lists/description/

描述

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.

输入

输入两个不带环的单链表。

输出

两个单链表的交集。

算法思想：

首先初始化让指针pa和pb分别指向单链表的表头，循环直到pa==pb，如果不相等，则判断pa是否为空，如不为空，则pa=pa->next,如果为空则pa=headB，同理，pb也是如此。循环完之后，如果两个有交集则会返回交集的起始位置，否则将会返回空。这里在pa或者pb为空时，需要将其指向另一个单表链的表头，这样才能保证遍历的总长度相同，如果两个单链表的长度不同且有交集则遍历的长度为O（n + m – k）,其中n、m为两个单链表的长度，k为交集的长度，如果两个单链表的交集为空，则遍历的长度为O（n + m）。

源代码

/** * 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 *pa = headA,*pb =headB;
        while(pa != pb)
        {
            pa = pa?pa->next:headB;
            pb = pb?pb->next:headA;
        }
        return pa;
    }
};
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