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
.
链表分区,把比x小的放在前面,其余的放在后面,并不改变其相对位置。
思路是建立两个链表分别存放大数和小数,再把两个链表合在一起
/**
* 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||head.next==null)return head;
ListNode shead = new ListNode(0);
ListNode lhead = new ListNode(0);
ListNode s = shead;
ListNode l = lhead;
while(head!=null){
if(head.val>=x){
l.next = head;
l = l.next;
}
else{
s.next = head;
s = s.next;
}
head = head.next;
}
s.next = lhead.next;
l.next = null;
return shead.next;
}
}