21. Partition List

Given the head of 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.

Example 1:

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

Input: head = [2,1], x = 2
Output: [1,2]

Solution: (Using Additional Vector and two traversal)

Solution: (Two pointers)

class Solution
{
public:
    ListNode *partition(ListNode *head, int x)
    {

        ListNode *sm = new ListNode;
        ListNode *s = sm;
        sm->next = NULL;
        sm->val = -5000;

        ListNode *gr = new ListNode;
        ListNode *g = gr;
        gr->next = NULL;
        gr->val = 5000;

        ListNode *p = head;

        while (p)
        {
            if (p->val < x)
            {
                s->next = p;
                s = s->next;
            }
            else if (p->val >= x)
            {
                g->next = p;
                g = g->next;
            }

            p = p->next;
        }

        g->next = NULL;

        if (gr->next)
        {
            s->next = gr->next;
        }

        return sm->next;
    }
};

Time Complexity: O(n) Space Complexity: O(1)