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.
创建两个链表，一个比x小，一个大于等于x。最后把它们连起来。注意把第二个链表最后一个元素置为NULL。不然可能产生死循环。
1
class
Solution {
2
public
:
3
ListNode *partition(ListNode *head,
int
x) {
4
ListNode n1(
0
), n2(
0
);
5
ListNode *p1 = &n1, *p2 = &
n2;
6
7
while
(head !=
NULL) {
8
if
(head->val <
x) {
9
p1->next =
head;
10
p1 = p1->
next;
11
}
else
{
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p2->next =
head;
13
p2 = p2->
next;
14
}
15
head = head->
next;
16
}
17
p2->next =
NULL;
18
p1->next =
n2.next;
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return
n1.next;
20
}
21
};
这里n1和n2不用new，省得最后还要delete.
