86. Partition List

• Difficulty: Medium

• Topics: Linked List, Two Pointers

• Similar Questions:

Problem:

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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

Solutions:

/**
 * 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* left = new ListNode(0);
        ListNode* right = new ListNode(0);

        ListNode* leftTail = left;
        ListNode* rightTail = right;

        while (head) {
            if (head->val < x) {
                leftTail->next = head;
                leftTail = head;
            } else {
                rightTail->next = head;
                rightTail = head;
            }

            head = head->next;
        }

        leftTail->next = right->next;
        rightTail->next = nullptr;
        return left->next;
    }
};