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Partition List

题目

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.

方法

将原来的Linked List 拆分成两个, 一个记录比X小的元素,另一个记录比X大的元素,最后将其合并。
    public ListNode partition(ListNode head, int x) {
        ListNode lessStart = null;
        ListNode lessEnd = null;
        ListNode greaterStart = null;
        ListNode greaterEnd = null;
        ListNode cur = head;
        while (cur != null) {
            if (cur.val < x) {
                if (lessStart == null) {
                    lessStart = cur;
                    lessEnd = cur;
                } else {
                    lessEnd.next = cur;
                    lessEnd = cur;
                }
            } else {
                if (greaterStart == null) {
                    greaterStart = cur;
                    greaterEnd = cur;
                } else {
                    greaterEnd.next = cur;
                    greaterEnd = cur;
                }
            }
            cur = cur.next;
        }
        if (lessStart == null) {
            return greaterStart;
        }
        if (greaterStart == null) {
            return lessStart;
        }
        greaterEnd.next = null;
        lessEnd.next = greaterStart;
        return lessStart;
    }