LeetCode160 | Intersection-of-two-linked-lists

题目地址(160. 相交链表)

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

题目描述

编写一个程序，找到两个单链表相交的起始节点。如下面的两个链表：在节点 c1 开始相交。示例 1：输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
输出：Reference of the node with value = 8
输入解释：相交节点的值为 8 （注意，如果两个链表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [4,1,8,4,5]，链表 B 为 [5,0,1,8,4,5]。在 A 中，相交节点前有 2 个节点；在 B 中，相交节点前有 3 个节点。示例 2：输入：intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
输出：Reference of the node with value = 2
输入解释：相交节点的值为 2 （注意，如果两个链表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [0,9,1,2,4]，链表 B 为 [3,2,4]。在 A 中，相交节点前有 3 个节点；在 B 中，相交节点前有 1 个节点。示例 3：输入：intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
输出：null
输入解释：从各自的表头开始算起，链表 A 为 [2,6,4]，链表 B 为 [1,5]。由于这两个链表不相交，所以 intersectVal 必须为 0，而 skipA 和 skipB 可以是任意值。
解释：这两个链表不相交，因此返回 null。注意：如果两个链表没有交点，返回 null.
在返回结果后，两个链表仍须保持原有的结构。
可假定整个链表结构中没有循环。
程序尽量满足 O(n) 时间复杂度，且仅用 O(1) 内存。

前置知识

本题中单链表ListNode数据结构包括一个整型的val，和一个指针next，这意味着对于给定的两条单链表要么无重复部分，要么有重复部分但绝对是“分-合”的状态，绝不会是“分-合-分”。

思路

对单链表A和B的值逐一对应比较，当A（这里假设A比较短）走到A的尾部后，若仍没有结束则转向B的头部从B开始；同理B（这里假设B比较长）走到B的尾部后，转向A的头部从A开始。

关键点

这样做其实表示l1与l2走过的距离是相同的，两者如果有intersect则必定最快得到结果
A(a+c),B(b+c) ====>>>> a + c + b = b + c + a

代码

语言支持：Java

Java Code:


/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {// ListNode是函数的返回值，即对应节点的valuepublic ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode l1 = headA, l2 = headB;while (l1 != l2) {l1 = (l1 == null) ? headB : l1.next;l2 = (l2 == null) ? headA : l2.next;}return l1;}
}

复杂度分析

令 n 为数组长度。

时间复杂度：O(a+b)O(a+b)O(a+b)
空间复杂度：O(1)O(1)O(1)

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LeetCode160 | Intersection-of-two-linked-lists

题目地址(160. 相交链表)

题目描述

前置知识

思路

关键点

代码

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