[leetcode: Python]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 may not modify the values in the list, only nodes itself can be changed.

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方法一：49ms
交换的四个步骤如下图
这里加了个头结点。

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def swapPairs(self, head):        """        :type head: ListNode        :rtype: ListNode        """        if head is None or head.next is None:            return head        dummy = ListNode(0)        dummy.next = head        p = dummy        while p.next and p.next.next:            tmp = p.next.next            p.next.next = tmp.next            tmp.next = p.next            p.next = tmp            p = p.next.next        return dummy.next

方法二：38ms

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def swapPairs(self, head):        """        :type head: ListNode        :rtype: ListNode        """        pre = self        pre.next = head        while pre.next and pre.next.next :            a = pre.next            b = a.next            a.next = b.next            pre.next = b            b.next = a            pre = a        return self.next

方法三：35ms
递归

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def swapPairs(self, head):        """        :type head: ListNode        :rtype: ListNode        """        if not head or not head.next: return head        cur = head        head = cur.next        cur.next = self.swapPairs(head.next)        head.next = cur        return head        '''        if not head or not head.next: return head        dummy = ListNode(0)        dummy.next = head        cur = dummy        while cur.next and cur.next.next:            # 如果不方便检查空，把ref挪到while里面让while来查也是很好的            n1 = cur.next            n2 = cur.next.next            n1.next = n2.next            n2.next = n1            cur.next = n2            cur = n1        return dummy.next        '''