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：

用两个指针less和greater分别保存小于x和大于x的列表，然后遍历一遍，一边遍历一边更新这两个指针。

代码：

    ListNode *partition(ListNode *head, int x) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == NULL)
            return NULL;
        ListNode *less=NULL, *greater=NULL;
        ListNode *newhead = head, *greaterhead = NULL;
        while(head){
            ListNode *tmp = head->next;
            if(head->val < x){
                if(less == NULL){
                    less = head;
                    newhead = less;
                }    
                else{
                    head->next = less->next;
                    less->next = head;
                    less = less->next;
                }
            }
            else{
                if(greater == NULL){
                    greaterhead = head;
                    greater = head;
                    greater->next = NULL;
                }
                else{
                    greater->next = head;
                    greater = greater->next;
                    greater->next = NULL;
                }
            }
            head = tmp;
        }
        if(less)
            less->next = greaterhead;
        return newhead;
    }