Difficulty:: Medium
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:
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| Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5
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Solution
Language: Java
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class Solution { public ListNode partition(ListNode head, int x) { if (head == null) { return null; } ListNode less = new ListNode(0); ListNode lessEnd = less; ListNode large = new ListNode(0); ListNode largeEnd = large; while (head != null) { if (head.val < x) { lessEnd.next = head; lessEnd = head; } else { largeEnd.next = head; largeEnd = head; } head = head.next; } lessEnd.next = large.next; largeEnd.next = null; return less.next; } }
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