{
   "question_id": 24,
   "question_title": "两两交换链表中的节点",
   "difficulty": "中等",
   "question_content": "<p>给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。</p><p><strong>你不能只是单纯的改变节点内部的值</strong>，而是需要实际的进行节点交换。</p><p> </p><p><strong>示例 1：</strong></p><img alt=\"\" src=\"https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0024.Swap%20Nodes%20in%20Pairs/images/swap_ex1.jpg\" style=\"width: 422px; height: 222px;\" /><pre><strong>输入：</strong>head = [1,2,3,4]<strong><br />输出：</strong>[2,1,4,3]</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>head = []<strong><br />输出：</strong>[]</pre><p><strong>示例 3：</strong></p><pre><strong>输入：</strong>head = [1]<strong><br />输出：</strong>[1]</pre><p> </p><p><strong>提示：</strong></p><ul>\t<li>链表中节点的数目在范围 <code>[0, 100]</code> 内</li>\t<li><code>0 <= Node.val <= 100</code></li></ul><p> </p><p><strong>进阶：</strong>你能在不修改链表节点值的情况下解决这个问题吗?（也就是说，仅修改节点本身。）</p>",
   "topic_link": "https://bbs.csdn.net/topics/600470117",
   "cpp": "#include <bits/stdc++.h>\nusing namespace std;\nstruct ListNode\n{\n\tint val;\n\tListNode *next;\n\tListNode() : val(0), next(nullptr) {}\n\tListNode(int x) : val(x), next(nullptr) {}\n\tListNode(int x, ListNode *next) : val(x), next(next) {}\n};\nclass Solution\n{\npublic:\n\tListNode *swapPairs(ListNode *head)\n\t{\n\t\tstruct ListNode dummy, *prev = &dummy, *p = head;\n\t\tdummy.next = head;\n\t\twhile (p != nullptr && p->next != nullptr)\n\t\t{\n\t\t\tstruct ListNode *q = p->next;\n\t\t\tp->next = q->next;\n\t\t\tq->next = prev->next;\n\t\t\tprev->next = q;\n\t\t\tprev = p;\n\t\t\tp = p->next;\n\t\t}\n\t\treturn dummy.next;\n\t}\n};",
   "java": "public class ListNode {\n\tint val;\n\tListNode next;\n\tListNode(int x) {\n\t\tval = x;\n\t}\n}\nclass Solution {\n\tpublic ListNode swapPairs(ListNode head) {\n\t\tListNode list1 = new ListNode(0);\n\t\tlist1.next = head;\n\t\tListNode list2 = list1;\n\t\twhile (head != null && head.next != null) {\n\t\t\tlist2.next = head.next;\n\t\t\thead.next = list2.next.next;\n\t\t\tlist2.next.next = head;\n\t\t\tlist2 = list2.next.next;\n\t\t\thead = list2.next;\n\t\t}\n\t\treturn list1.next;\n\t}\n}",
   "js": "",
   "python": "class ListNode(object):\n\tdef __init__(self, x):\n\t\tself.val = x\n\t\tself.next = None\nclass LinkList:\n\tdef __init__(self):\n\t\tself.head=None\n\tdef initList(self, data):\n\t\tself.head = ListNode(data[0])\n\t\tr=self.head\n\t\tp = self.head\n\t\tfor i in data[1:]:\n\t\t\tnode = ListNode(i)\n\t\t\tp.next = node\n\t\t\tp = p.next\n\t\treturn r\n\tdef\tconvert_list(self,head):\n\t\tret = []\n\t\tif head == None:\n\t\t\treturn\n\t\tnode = head\n\t\twhile node != None:\n\t\t\tret.append(node.val)\n\t\t\tnode = node.next\n\t\treturn ret\nclass Solution(object):\n\tdef swapPairs(self, head):\n\t\tdummyHead = ListNode(-1)\n\t\tdummyHead.next = head\n\t\tprev, p = dummyHead, head\n\t\twhile p != None and p.next != None:\n\t\t\tq, r = p.next, p.next.next\n\t\t\tprev.next = q\n\t\t\tq.next = p\n\t\t\tp.next = r\n\t\t\tprev = p\n\t\t\tp = r\n\t\treturn dummyHead.next\n# %%\nl = LinkList()\nhead = [1,2,3,4]\nl1 = l.initList(head)\ns = Solution()\nprint(l.convert_list(s.swapPairs(l1)))",
   "status": 1,
   "keywords": "递归,链表",
   "license": {
      "cpp": "https://github.com/begeekmyfriend/leetcode",
      "python": "https://github.com/qiyuangong/leetcode",
      "java": "https://blog.csdn.net/FaustoPatton/article/details/82993032"
   },
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