leetcode : 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.

简单题，遍历数组，将小于x的节点插入到less链表，大于x的插入到greater链表，然后将两个链表连接即可。

注意把链表最后一个节点的next指针修改为NULL，防止出现环

代码:

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode *less = NULL,*greater = NULL;
        ListNode *p1 = NULL, *p2 = NULL;
        ListNode *h = head;
        while(h){
            if(h->val < x){
                if(!less){
                    less = h;
                }else{
                    p1->next = h;
                }
                p1 = h;
            }else{
                if(!greater){
                    greater = h;
                }else{
                    p2->next = h;
                }
                p2 = h;
            }
            h = h->next;
        }
        if(less){　　　　　　　　　　　　　　　　　　//连接两个链表
            head = less;
            if(p1)  p1->next = greater;
        }else
            head = greater;
        if(p2)　　　　　　　　　　　　　　　　　　//防止出现环
            p2->next = NULL;
        return head;
    }
};