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

.

思路：此题的任务是划分链表，比x小的链表节点都出现在比x大或者等于的节点前面，而且还要保持原来的相对位置不要发生变化。我们新建两个链表了l1,l2，遍历原来的链表 ，把值小于x的节点都加入l1,把值大于或者等于x的节点都加入l2，最后将l2连接至l1的后面，所得的链表就是我们所求的结果了。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode *head1,*head2;
head1=head2=NULL;
ListNode *cur1,*cur2;
cur1=cur2=NULL;
while(head){
ListNode *cur=head;
head=head->next;
cur->next=NULL;
if(cur->val<x){
if(!head1){
head1=cur;
cur1=head1;
}
else{
cur1->next=cur;
cur1=cur1->next;
}

}else{
if(!head2){
head2=cur;
cur2=head2;
}else{
cur2->next=cur;
cur2=cur2->next;
}
}
}
if(!cur1)
return head2;
cur1->next=head2;
return head1;

}
};