Leetcode 143. Reorder List

Description: Given a singly linked list L: L₀→L₁→…→L_n-1→L_n, reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-2→… You may not modify the values in the list's nodes, only nodes itself may be changed.

Link: https://leetcode.com/problems/reorder-list/

Examples:

Example 1:
Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

思路: 最直接的做法就是每次将链表分成两部分，reordered part and the part to reorder, 已经重新排序过的部分和待排序的部分，记录 the tail of reordered part is pointer p and the head of the part to reorder is q. 每次 1) 找到待排序链表的结尾，将尾节点的前一个node记为pre，pre.next = None; 2) 将尾节点连接到p, 将p.next连接到尾节点; 3) update p and q. 直到p.next = None (即没有待排序的部分)，或者q.next = None (待排序的部分只有一个点)。

class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """
        if head and head.next and head.next.next:
            p = head
            q = p.next
            while p.next and q.next:
                # print(p.val)
                while q.next:
                    pre = q
                    q = q.next
                pre.next = None
                s = p.next
                p.next = q
                q.next = s
                p = s
                q = p.next

但这样会超时的，因为每次都要找待排序的尾节点。为了不要每次查找，就需要存储。第一次遍历时，记录所有的节点并计算总长度length, 再次遍历时是需要遍历总链表长度的一半，步骤和上面是一样的，只是待排序的尾节点不需要遍历到结尾才能找到，而是直接在nodes list中根据位置 i 找到。这样时间和空间都很好。

class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """
        p = head
        length = 0
        nodes = list()
        while p:
            nodes.append(p)
            length += 1
            p = p.next
        if length % 2 == 0:
            N = int(length/2)
        else:
            N = int(length/2) + 1
#         print(length, N)
        i = 0
        while i < N-1:
            p = nodes[i]  # update p
            q = nodes[length-i-1]  # find tail of q
            nodes[length-i-2].next = None # pre.next = None
            s = p.next
            p.next = q
            q.next = s
            i += 1         

日期: 2020-11-27  快周末了呢， 激动的心！