LeetCode---(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

.

题意为将小于特定值x的所有节点均放在不小于x的节点的前面，而且，不能改变原来节点的前后位置。

思路：设置两个链表，一个用于存放值小于x的节点，一个用于存放值不小于x的节点。

/**
* 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) {
if(head==NULL)
return NULL;
ListNode* p=head;
ListNode* L1=new ListNode(0);
ListNode* p1=L1;
ListNode* L2=new ListNode(0);
ListNode* p2=L2;
while(p!=NULL)
{
if(p->val<x)
{
ListNode* q=new ListNode(p->val);
p1->next=q;
p1=p1->next;
}
else
{
ListNode* q=new ListNode(p->val);
p2->next=q;

99b3
p2=p2->next;
}
p=p->next;
}
p1->next=L2->next;
return L1->next;
}
};