LeetCode 24. Swap Nodes in Pairs 解题报告

24. Swap Nodes in Pairs

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Total Accepted: 82451 Total Submissions: 239340 Difficulty: Medium

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

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    一道链表的题目，考验对指针的理解。大四那年去小米三面的时候KK考了我两道题，其中一道就是这个，至今记忆犹新。今天重做一遍。虽然当时KK肯定了我的code，但如今看来，还是有小bug的。。。。。

    我的AC代码

public class SwapNodesinPairs {/** * @param args */public static void main(String[] args) {ListNode h = new ListNode(1);ListNode h2 = new ListNode(2);ListNode h3 = new ListNode(3);ListNode h4 = new ListNode(4);h.next = h2;h2.next = h3;h3.next = h4;System.out.print(swapPairs(h));}public static ListNode swapPairs(ListNode head) {if (head == null || head.next == null)return head;ListNode h = new ListNode(0);h.next = head;ListNode pre = h, p1 = head, p2 = head.next;while (p1 != null && p2 != null) {// swappre.next = p2;p1.next = p2.next;p2.next = p1;// 指针后移pre = p1;p1 = p1.next;if (p1 != null)p2 = p1.next;}return h.next;}}