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:

/*** 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;ListNode curA = headA,curB = headB;while (curA!=null&&curB!=null){if (curA.next ==null&&curB.next==null&&curA!=curB)return null; //如果走到了鏈的盡頭，而這盡頭又不是交//點，說明兩條鏈不相交else if (curA==curB) //如果同時到交點了，就跳出循環break;if (curA.next == null) //下面這段可能需要畫圖理解curA = headB; //如果兩條鏈有交點，則curA和curB到else //交點的距離是相同的curA = curA.next;if (curB.next == null)curB = headA;else curB = curB.next;}return curA;} }

?

• • 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.

轉載于:https://www.cnblogs.com/David-Lin/p/7778509.html

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