[LeetCode] Challenge log 143

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143. Reorder List

Algorithm:

  1. split the linklist
  2. reverse the second half
  3. merge one by one

Given a singly linked list L: L0→L1→…→L**n-1→Ln,
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.

Example 1:

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

Example 2:

1
Given 1->2->3->4->5, reorder it to 1->5->2->4->3.
Soulution:
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class Solution:
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: void Do not return anything, modify head in-place instead.
        """
        if not head or not head.next: 
            pass
        else:
            # return node at len // 2    
            newHead = ListNode(0)
            newHead.next = head
            slow = newHead
            fast = newHead
            while fast.next and fast.next.next:
                fast = fast.next.next
                slow = slow.next
            mid = slow

            # reverse the second half as head2
            tail = None
            cur = mid.next
            mid.next = None # important! aviod cycle!!!
            while cur:
                nxt = cur.next
                cur.next = tail
                tail = cur
                cur = nxt
            head2 = tail


            # link one by one
            p1 = head
            p2 = head2
            res = ListNode(0)
            cur = res
            while p1 or p2:
                if p1: 
                    cur.next = p1
                    cur = cur.next
                    p1 = p1.next
                if p2:
                    cur.next = p2
                    cur = cur.next
                    p2 = p2.next
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