[Leetcode] 24. Swap Nodes in Pairs

原创于 2017-02-27 21:28:05 发布 · 204 阅读

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本文介绍了一种在常数空间复杂度下交换链表中每两个相邻节点的算法。通过迭代方式实现，每次操作记录前一节点并更新指针指向，最终返回修改后的链表头。该方法不改变节点值，仅调整节点位置。

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Problem:

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.

Idea:
Just do swap operation every two nodes. Therefore, after every swap operation, we need to move tmpNode backwards by two Nodes. Kindly note that the previous node should be recorded for the following swap operation.

Solution:

class Solution(object):
    def swapPairs(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        preheadNode = ListNode(-1)
        preheadNode.next = head
        tmpNode = None if head == None else head.next
        preNode = preheadNode
        while tmpNode != None:
            nextNode = tmpNode.next
            tmpNode.next = preNode.next
            preNode.next.next = nextNode
            preNode.next = tmpNode
            preNode = tmpNode.next
            tmpNode = None if nextNode == None else nextNode.next
        return preheadNode.next