leetcode解题之24. Swap Nodes in Pairs java 版（成对翻转链表）

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24. Swap Nodes in Pairs

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 maynot modify the values in the list, only nodes itself can be change

成对交换链表

自己的笨方法，AC，准备两个list，一个存放第奇数个结点，一个存放偶数个结点，取的时候反过来。合成一个，最后就是结果，注意最后需要把尾结点置空

public ListNode swapPairs(ListNode head) {if (head == null)return null;List<ListNode> first = new ArrayList<>();List<ListNode> second = new ArrayList<>();ListNode cur = head;ListNode ph = new ListNode(0);int i = 0;while (cur != null) {if ((i & 1) == 0)first.add(cur);elsesecond.add(cur);cur = cur.next;i++;}cur = ph;while (!first.isEmpty() && !second.isEmpty()) {cur.next = second.get(0);cur = cur.next;cur.next = first.get(0);cur = cur.next;first.remove(0);second.remove(0);}if (!second.isEmpty()) {cur.next = second.get(0);cur = cur.next;}if (!first.isEmpty()) {cur.next = first.get(0);cur = cur.next;}cur.next = null;return ph.next;}

参考代码：

public ListNode swapPairs(ListNode head) {if (head == null || head.next == null)return head;// 防止丢链ListNode tmpHead = new ListNode(0);// ptr1始终指向需要交换的pair的前面一个node，ListNode ptr1 = tmpHead;// ptr2始终指向需要交换的pair的第一个node。ListNode ptr2 = head;// 当链表长度为奇数时，ptr2.next可能为null；// 当链表长度为偶数时，ptr2可能为null。while (ptr2 != null && ptr2.next != null) {// 需要用一个临时指针nextstart， 指向下一个，// 需要交换的pair的第一个node保证下一次交换的正确进行。ListNode nextstart = ptr2.next.next;ptr2.next.next = ptr2;ptr1.next = ptr2.next;ptr2.next = nextstart;ptr1 = ptr2;ptr2 = nextstart;}return tmpHead.next;}

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