86. Partition List

题目描述

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given1->4->3->2->5->2andx= 3,
return1->2->2->4->3->5.

解题方法

1. 因為linked list 的特性，swap太麻煩，故不考慮
2. 利用dummy left /right to merge

    public ListNode partition(ListNode head, int x) {
        if (head == null) {
            return null;
        }

        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy, 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;
        left.next = rightDummy.next;
        return leftDummy.next;
    }