Partition List

Lintcode Online Judge

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->null and x = 3,
return 1->2->2->4->3->5->null.

    public ListNode partition(ListNode head, int x) {
        ListNode leftDummy = new ListNode(0);
        ListNode left = leftDummy;
        ListNode rightDummy = new ListNode(0);
        ListNode right = rightDummy;
        while(head!=null){
            if(head.val<x){
                left.next = head;
                left = left.next;
            }else{
                right.next = head;
                right = right.next;
            }
            head = head.next;
        }
        right.next = null; //important
        left.next = rightDummy.next;
        
        return leftDummy.next;
    }
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