| 标题： | Partition List |
| 通过率： | 27.5% |
| 难度： | 中等 |
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
.
本题就是利用一个target将链表分成两部分，小于target和大于target的
用两个链表进行记录，small和big进行连接时候一定要将.next置null，否则内存会一直叠加。连接的过程就是指针的重定向过程，所以一定要置null，要不然会讲整个链表连接过来，代码如下：
1
/**
2
* Definition for singly-linked list.
3
* public class ListNode {
4
* int val;
5
* ListNode next;
6
* ListNode(int x) {
7
* val = x;
8
* next = null;
9
* }
10
* }
11
*/
12
public
class
Solution {
13
public
ListNode partition(ListNode head,
int
x) {
14
ListNode small=
new
ListNode(1
);
15
ListNode big=
new
ListNode(1
);
16
ListNode s=small,b=
big;
17
if
(head==
null
)
return
head;
18
while
(head!=
null
){
19
if
(head.val<
x){
20
s.next=
head;
21
head=
head.next;
22
s=
s.next;
23
s.next=
null
;
24
}
25
else
{
26
b.next=
head;
27
head=
head.next;
28
b=
b.next;
29
b.next=
null
;
30
}
31
}
32
s.next=
big.next;
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return
small.next;
34
}
35
}
