Question

Formatted question description: https://leetcode.ca/all/86.html

86	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.

@tag-linkedlist

Algorithm

All nodes less than the given value are taken out to form a new linked list. At this time, the values of the remaining nodes in the original linked list are greater than or equal to the given value, as long as the original linked list is directly connected to the new linked list.

Original: 1 -> 4 -> 3 -> 2 -> 5 -> 2
New:

Original: 4 -> 3 -> 2 -> 5 -> 2
New:　  1

Original: 4 -> 3 -> 5 -> 2
New:　  1 -> 2

Original: 4 -> 3 -> 5
New:　  1 -> 2 -> 2

Original:
New:　  1 -> 2 -> 2 -> 4 -> 3 -> 5

Code

Java

public class Partition_List {

	/**
	 * 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 head;
	        }

	        ListNode lessList = new ListNode(0);
	        ListNode moreList = new ListNode(0);

	        ListNode lessHeadCopy = lessList;
	        ListNode moreHeadCopy = moreList;

	        int xCount = 0;
	        ListNode current = head;
	        while (current != null) {
	            int num = current.val;

	            if (num < x) {
	                lessList.next = new ListNode(num);
	                lessList = lessList.next;
	            } else {
	                moreList.next = new ListNode(num);
	                moreList = moreList.next;
	            }

	            current = current.next;
	        }

	        // append target x
	        while (xCount > 0) {
	            lessList.next = new ListNode(x);
	            lessList = lessList.next;

	            xCount--;
	        }

	        // connect two lists
	        lessList.next = moreHeadCopy.next;

	        return lessHeadCopy.next;
	    }
	}

}

// OJ: https://leetcode.com/problems/partition-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode ltHead, geHead, *ltTail = &ltHead, *geTail = &geHead;
        while (head) {
            auto p = head;
            head = head->next;
            if (p->val < x) {
                ltTail->next = p;
                ltTail = p;
            } else {
                geTail->next = p;
                geTail = p;
            }
        }
        ltTail->next = geHead.next;
        geTail->next = NULL;
        return ltHead.next;
    }
};

# 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
    """
    if head is None:
      return None
    dummy = ListNode(-1)
    dummy.next = head
    sHead = sDummy = ListNode(-1)
    p = dummy
    while p and p.next:
      if p.next.val < x:
        sDummy.next = p.next
        p.next = p.next.next
        sDummy = sDummy.next
      else:
        p = p.next
      # if you change p.next then make sure you wouldn't change p in next run
    sDummy.next = dummy.next
    return sHead.next

86 - Partition List

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