Leetcode: Partition List

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
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