LeetCode 143 [Reorder List]

原题

给定一个单链表L: L0→L1→…→Ln-1→Ln,
重新排列后为：L0→Ln→L1→Ln-1→L2→Ln-2→…
必须在不改变节点值的情况下进行原地操作。

样例
给出链表 1->2->3->4->null，重新排列后为1->4->2->3->null。

解题思路

• 把题目分解成一个个小的任务
• 首先根据1->2->3->4，找中点，然后得到两个链表1->2和3->4
• 翻转第二个链表得到1->2和4->3
• 最后合并两个链表得到1->4->2->3

完整代码

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: void Do not return anything, modify head in-place instead.
        """
        if head == None or head.next == None:
            return
        
        mid = self.findMiddle(head)
        tail = self.reverse(mid.next)
        mid.next = None
        
        self.merge(head, tail)
        
    def reverse(self, head):
        newHead = None
        while head != None:
            temp = head.next
            head.next = newHead
            newHead = head
            head = temp
        return newHead
        
    def merge(self, head1, head2):
        index = 0
        dummy = ListNode(0)
        while head1 != None and head2 != None:
            if index % 2 == 0:
                dummy.next = head1
                head1 = head1.next
            else:
                dummy.next = head2
                head2 = head2.next
            index += 1
            dummy = dummy.next
        if head1:
            dummy.next = head1
        if head2:
            dummy.next = head2
    
    def findMiddle(self, head):
        slow, fast = head, head.next
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow