86

86. Partition List (Medium)

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]

 

Constraints:

• The number of nodes in the list is in the range [0, 200].
• -100 <= Node.val <= 100
• -200 <= x <= 200

Related Topics:
Linked List, Two Pointers

Similar Questions:

• Partition Array According to Given Pivot (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/partition-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode ltHead, *ltTail = &ltHead, geHead, *geTail = &geHead;
        while (head) {
            auto n = head;
            head = head->next;
            if (n->val < x) {
                ltTail->next = n;
                ltTail = n;
            } else {
                geTail->next = n;
                geTail = n;
            }
        }
        ltTail->next = geHead.next;
        geTail->next = nullptr;
        return ltHead.next;
    }
};