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,
Given1->4->3->2->5->2and x = 3,
return1->2->2->4->3->5.
/**
* 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)
return null;
ListNode smallHead=new ListNode(0);
ListNode bigHead=new ListNode(0);
ListNode small=smallHead;
ListNode big=bigHead;
ListNode p=head;
while(p!=null){
if(p.val>=x){
big.next=p;
big=big.next;
}else{
small.next=p;
small=small.next;
}
p=p.next;
}
small.next=bigHead.next;
big.next=null;//必须把后面断掉，因为原来的节点还指着后面。我这里犯错误了，没有断掉，导致报错
return smallHead.next;
}
}