Leetcode-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

.

Analysis:

The first element that is not less than x is the point of partition. We record this node and its predecessor. Once we encount a node that is less than x later, we insert it between the point and its predecessor, then update the predecssor to this node.

Solution:

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
if (head==null || head.next==null) return head;

ListNode preHead = new ListNode(0);
preHead.next = head;
ListNode end = preHead;
ListNode cur = head;

while (cur!=null && cur.val<x){
end = cur;
cur = cur.next;
}

while (cur!=null && cur.next!=null){
if (cur.next.val>=x){
cur = cur.next;
} else {
ListNode temp = cur.next;
cur.next = temp.next;
temp.next = end.next;
end.next = temp;
end = temp;
}
}

return preHead.next;
}
}