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.

递归的解法

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None or head.next ==None:
            return head
        temp = head.next
        head.next = self.swapPairs(temp.next)
        temp.next = head
        return temp
直接循环的解法
class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None or head.next ==None:
            return head
        p1 = head
        p2 = head.next
        p1.next = p2.next
        p2.next=p1
        result = p2
        while True:
            p0 = p1
            p1 = p1.next
            if p1==None or p1.next ==None:
                return result
            else :
                p2 = p1.next
                p1.next = p2.next
                p2.next = p1
                p0.next = p2
 
原文地址:https://www.cnblogs.com/feiqiangs/p/5792475.html