動機

真的看不懂以前寫什麼

Problem

Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.

Note that the relative order inside both the even and odd groups should remain as it was in the input.

You must solve the problem in O(1) extra space complexity and O(n) time complexity.

 

Example 1:

Input: head = [1,2,3,4,5]Output: [1,3,5,2,4]

Example 2:

Input: head = [2,1,3,5,6,4,7]Output: [2,3,6,7,1,5,4]

 

Constraints:

  • n == number of nodes in the linked list
  • 0 <= n <= 104
  • -106 <= Node.val <= 106

Sol

善用變數去抽象

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head:
            return None
        # head 
        odd, even = head, head.next
        
        # collect odds, evens
        optr, eptr = odd, even
        oprev, eprev = None, None
        while optr and eptr and eptr.next: # 可以是None的前面要保證是有效的!!
            onext, enext = eptr.next, eptr.next.next # 可以是None的前面要保證是有效的!!
            eptr.next, optr.next = enext, onext
            oprev, eprev = optr, eptr
            optr, eptr = optr.next, eptr.next
        
        # break loop
        for p in [optr, eptr]:
            if p:
                p.next = None
        
        # connect
        if optr:
            optr.next = even
        else:
            oprev.next = even
        
        return odd