[LeetCode] Intersection of Two Linked Lists, twosumleetcode

[LeetCode] Intersection of Two Linked Lists, twosumleetcode

Question

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, returnnull.
• 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 shocould preferably run in O (n) time and use only O (1) memory.

Answer

The question requires that the first node of the intersection of two single-chain tables can be traversed by traversing the first linked list, calculating the length of len1, saving the last node, traversing the second linked list, and calculating the length of len2, at the same time check whether the last node is the same, and then traverse the two linked list from the beginning, (if len1> len2), the linked list first traverses the len1-len2 nodes, then traverse to the same node at the same time, the Code is as follows:

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode next; * ListNode (int x) {* val = x; * next = null; *} */public class Solution {public ListNode getIntersectionNode (ListNode headA, ListNode headB) {if (headA = null | headB = null) {return null ;} int len1 = 0; ListNode h1 = headA; while (h1.next! = Null) {// note that. next h1 = h1.next; len1 ++;} int len2 = 0; ListNode h2 = headB; while (h2.next! = Null) {h2 = h2.next; len2 ++;} ListNode n1 = headA; ListNode n2 = headB; if (len1> len2) {int k = len1-len2; while (k --> 0) {n1 = n1.next;} else {int k = len2-len1; while (k --> 0) {n2 = n2.next ;}} /* or while (n1! = N2) {n1 = n1.next; n2 = n2.next;} return n1; */while (n1! = Null & n2! = Null) {if (n1 = n2) {return n1;} n1 = n1.next; n2 = n2.next;} return null ;}}

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