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.

使用两个链表newHead1, newHead2，遍历原链表，如果结点值小于x，则挂在newHead1上，如果大于等于x，则挂在newHead2上，最后把newHead2挂在newHead1上。

/**
 * 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) {
        ListNode* newHead1 = new ListNode(0);
        ListNode* newHead2 = new ListNode(0);
        
        ListNode* pNode1 = newHead1;
        ListNode* pNode2 = newHead2;
        
        ListNode* pNode = head;
        
        while(pNode){
            if(pNode->val < x){
                pNode1->next = pNode;
                pNode1 = pNode1->next;
            }else{
                pNode2->next = pNode;
                pNode2 = pNode2->next;
            }
            pNode = pNode->next;
        }
        
        pNode2->next = NULL;
        pNode1->next = newHead2->next;
        
        return newHead1->next;
    }
};