題目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.
public class Solution {
public ListNode partition(ListNode head, int x) {
if(head == null || head.next == null){
return head;
}
ListNode lessHead = new ListNode(1);
ListNode lessTail = lessHead;
ListNode moreHead = new ListNode(1);
ListNode moreTail = moreHead;
ListNode node = head;
ListNode tempNode = null;
while(node != null){
tempNode = node.next;
if(node.val < x){
lessTail.next = node;
lessTail = lessTail.next;
}else{
moreTail.next = node;
moreTail = moreTail.next;
}
node = tempNode;
}
lessTail.next = moreHead.next;
moreTail.next = null;
return lessHead.next;
}
}
