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.

题目解析：

方法类似快速排序。不过这类要设置两个指针，一个指向小于x结点的链表指针small，一个指向大于等于x的链表指针big。在设置两个指针psmall和pbig分别指向上述链表的尾结点。

有个需要注意的地方，遍历完整个链表后，pbig->next可能不为空，如果直接让psmall = big，可能造成环。所以每次区分完一个结点后，都将pbig->next = NULL; 由于要这样，head = head->next就必须在其之前赋值，否则会造成链表断裂。

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        if(head == NULL || head->next == NULL)
            return head;
        ListNode *small=NULL,*big=NULL;
        ListNode *psmall,*pbig;
        int count = 0;
        while(head){
            if(head->val < x){
                if(small == NULL){
                    small = head;
                    psmall = small;
                }else{
                    psmall->next = head;
                    psmall = head;
                }
                head = head->next;
            }else{
                if(big == NULL){
                    big = head;
                    pbig = big;
                    head = head->next;
                }else{
                    pbig->next = head;
                    pbig = head;
                    head = head->next;
                }
                pbig->next = NULL;
            }
        }
        if(small == NULL){
            return big;
        }
        psmall->next = big;
        return small;
    }
};