Leetcode 86. Partition List

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该博客介绍了如何实现一个算法，将给定的链表根据节点值进行分区，使得所有值小于x的节点位于值大于等于x的节点之前，同时保持原有相对顺序。算法使用两个辅助列表分别存储小于x和大于等于x的值，然后按顺序合并回原链表。

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Problem

Given the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Algorithm

Use two lists to save the values that number bigger than x and less than x, respectively. Then combine them in order.

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        if not head:
            return head
        
        l_less = []
        l_greq = []
        
        p = head
        while p:
            if p.val < x:
                l_less.append(p.val)
            else:
                l_greq.append(p.val)
            p = p.next
        
        p = head
        
        for v in l_less:
            p.val = v
            p = p.next
            
        for v in l_greq:
            p.val = v
            p = p.next
            
        return head