[LeetCode] [Partition List 2012-04-30]

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.

the key point of this problem is "DO NOT FORGET TO SET NULL".

 

/**
 * 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 == NULL)    return NULL;
        
        bool bBig=false;
        bool bSmall=false;
        ListNode *p=head;
        
        ListNode * pBig = NULL;
        ListNode * pSmall = NULL;
        ListNode *pHead = NULL;
        ListNode *pHead2 = NULL;
        while(p!=NULL)
        {
            int v = p->val;
            if(v >= x)
            {
                if(pBig==NULL)
                {
                    pBig = p;
                    pHead2 = p;
                }
                else
                {
                    pBig->next=p;
                    pBig = pBig->next;
                }
            }
            if(v < x)
            {
                if(pSmall==NULL)
                {
                    pSmall = p;
                    pHead = p;
                }
                else
                {
                    pSmall->next=p;
                    pSmall = pSmall->next;
                }
            }
            p=p->next;
        }
        !!!!!!!!//this is really import, or there will be a circle in linklist!!!!!!!!!!
        if(pSmall != NULL)
        {
            pSmall->next = NULL;
        }
        if(pBig != NULL)
        {
            pBig->next = NULL;
        }
        
        if(pHead != NULL)
        {
            pSmall->next = pHead2;
        }
        else
        {
            pHead = pHead2;
        }
        return pHead;
    }
};