Question Definition
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.
For example, Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.
Java Solution
public ListNode partition(ListNode head, int x) {
if(head == null || head.next == null) return head;
ListNode node = head;
ListNode pre = null;
ListNode rHead = null;
ListNode rPre = null;
while(node != null){
if(node.val < x){
if(head == null) head = node;
if(pre != null) pre.next = node;
pre = node;
} else {
if(head == node) head = null;
if(rHead == null) rHead = node;
if(rPre != null) rPre.next = node;
rPre = node;
}
node = node.next;
}
if(pre != null) pre.next = rHead;
else return rHead;
if(rPre != null) rPre.next = null;
return head;
}
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