Partition List

Given 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.

For example,

Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/partition-list ## Basic Ideas: 3 pointer. ## p_last_less: last node which is less than x ## p_last_greater: last node which is greater than x ## p: trasverse the list ## Complexity: Time O(n), Space O(1) # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def partition(self, head, x): """ :type head: ListNode :type x: int :rtype: ListNode """ dummy_less = ListNode(None) dummy_less.next = None dummy_greater = ListNode(None) dummy_greater.next = None p, p_last_less, p_last_greater = head, dummy_less, dummy_greater while p: if p.val < x: p_last_less.next = p p_last_less = p_last_less.next else: p_last_greater.next = p p_last_greater = p_last_greater.next p = p.next # end the list if p_last_less: p_last_less.next = None if p_last_greater: p_last_greater.next = None if p_last_less: p_last_less.next = dummy_greater.next return dummy_less.next else: return dummy_greater.next

Original URL: https://code.dennyzhang.com/partition-list

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