[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.

Analysis:

• Create left part of list for nodes less than x
• create right part of list for nodes greater than or equal to x
• concatenate them

class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        ListNode* dummyHead1 = new ListNode(0);
        ListNode* dummyHead2 = new ListNode(0);
        ListNode* p1 = dummyHead1;
        ListNode* p2 = dummyHead2;
        while(head != NULL)
        {
            if (head->val < x)
            {
                p1->next = head;
                p1 = p1->next;
            }
            else
            {
                p2->next = head;
                p2 = p2->next;
            }
            head = head->next;
        }
        
        p2->next = NULL;
        p1->next = dummyHead2->next;
        
        return dummyHead1->next;
    }
};