Partition a linked 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.
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode d1 = new ListNode();
ListNode d2 = new ListNode();
ListNode t1 = d1, t2 = d2;
while (head != null) {
if (head.val < x) {
t1.next = head;
t1 = t1.next;
} else {
t2.next = head;
t2 = t2.next;
}
head = head.next;
}
t1.next = d2.next;
t2.next = null;
return d1.next;
}
}Related Problems
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