Day 40, 86 Partition List

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.
---------------------------------------------------------------------

/**
 * 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 head;
        ListNode *leftHead = NULL, *rightHead = NULL, *end;
        ListNode ** before = &leftHead, **after = &rightHead;
        while (head != NULL) {
            if (head->val >= x) {
                *after = head;
                after = &(head->next);
            }else {
                *before = head;
                before = &(head->next);
            }
            // use end to break the new attached node from the original list
            end = head;
            head = head->next;
            end->next = NULL;
        }
        *before = rightHead; // connect two lists
        return leftHead;
    }
};

Update on Sep-19-2014 
Without double-pointer

/**
 * 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 *head1 = NULL, *head2 = NULL,*tail1 = NULL,*tail2 = NULL;
        
        while (head != NULL) {
            if (head->val < x) {
                if (head1 == NULL) {
                    head1 = head;
                    tail1 = head;
                }else {
                    tail1->next = head;
                    tail1 = tail1->next;
                }
            }else {
                
                if (head2 == NULL) {
                    head2 = head;
                    tail2 = head;
                }else {
                    tail2->next = head;
                    tail2 = tail2->next;
                }
            }
            head = head->next;
        }
        
        // merge
        if (tail1 != NULL) tail1->next = head2;
        if (tail2 != NULL) tail2->next = NULL;
        if (head1 == NULL) return head2;
        return head1;
    }
};