[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.

类似于qsort的partition的思想，用两根指针记录左右两边。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
            
        ListNode *pPre = NULL;
        ListNode *p = head;
        ListNode *q = head;
        
        while(q->next)
            q = q->next;
            
        ListNode *tail = q;
        
        while(p != q)
        {
            if (p->val >= x)
            {
                ListNode *pNext = p->next;
                if (head == p)
                    head = pNext;
                tail->next = p;
                p->next = NULL;
                tail = p;
                if (pPre)
                    pPre->next = pNext;
                p = pNext;
            }
            else
            {
                pPre = p;
                p = p->next;
            }
        }
        
        if (p->val >= x && q != tail)
        {
            ListNode *pNext = p->next;
            if (head == p)
                head = pNext;
            tail->next = p;
            p->next = NULL;
            if (pPre)
                pPre->next = pNext;
        }
        
        return head;       
    }
};