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.

分析：用sm、sm_tail、bg、bg_tail分别记录小于x的头、尾、大于等于x的头、尾。

用时：8ms

特别注意注释部分！！在给head->next指定新的位置之前不能将其置空，否则会出问题喔

/**
 * 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 || !head->next) return head;
        
        ListNode* sm = new ListNode(0);
        ListNode* sm_tail = sm;
        ListNode* bg = new ListNode(0);
        ListNode* bg_tail = bg;
        while(head){
            if(head->val < x){
                sm_tail->next = head;
                //head = head->next;
                sm_tail = sm_tail->next;
                head = head->next;
                sm_tail->next = NULL;
            }
            else{
                bg_tail->next = head;
                head = head->next;
                bg_tail = bg_tail->next;
                //head = head->next;
                bg_tail->next = NULL;
            }
            //head = head->next;
        }
        sm_tail->next = bg->next;
        return sm->next;
    }
};