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Leetcode[86] Partition List

06 Nov 2015

###Task 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.

###Python

# 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
        """
        leftS = ListNode(0)
        leftHead = leftS
        rightS = ListNode(0)
        rightHead = rightS
        if head == None or head.next == None:
            return head
        while head:
            if head.val < x:
                leftHead.next = ListNode(head.val)
                leftHead = leftHead.next
            else:
                rightHead.next = ListNode(head.val)
                rightHead = rightHead.next
            head = head.next
        leftHead.next = rightS.next
        
        return leftS.next

###Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) {
            return null;
        }
        
        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode leftPoint = leftDummy;
        ListNode rightPoint = rightDummy;
        while (head != null) {
            if (head.val < x) {
                leftPoint.next = head;
                leftPoint = leftPoint.next;
            } else {
                rightPoint.next = head;
                rightPoint = rightPoint.next;
            }
            head = head.next;
        }
        //remember to make rightPoint points to null
        rightPoint.next = null;
        leftPoint.next = rightDummy.next;
        
        return leftDummy.next;
    }
}

###Points

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