Leetcode # 86. Partition List

https://leetcode.com/problems/partition-list

Solution

ge_cur.next = cur
ge_cur = cur

→ ge_cur = ge_cur.next = cur
⇒ 1. ge_cur ← cur
2. ge_cur.next ≡ cur.next ← cur ⇒ loop
ge_cur.next = ge_cur = cur
⇒ 1. ge_cur.next ← cur
2. ge_cur ← cur

參考 多重賦值 (Multiple Assignment) 時的計算順序

Time Complexity: O(n), n := length of linked list
Space Complexity: O(1)
(The input and output generally do not count towards the space complexity.)

class Solution:
  def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
    ls_head, ge_head = ListNode(), ListNode()
    ls_cur, ge_cur, cur = ls_head, ge_head, head
    while cur:
      if cur.val >= x:
        ge_cur.next = ge_cur = cur
        cur = cur.next
        ge_cur.next = None
      else:
        ls_cur.next = ls_cur = cur
        cur = cur.next
        ls_cur.next = None
    ls_cur.next = ge_head.next
    return ls_head.next

Last Updated on 2023/08/16 by A1go