LeetCode 086 Partition List

题目

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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.

思路

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1 遍历原链表，根据大小分别放入制作成另外两个链表。

2 遍历完成后，注意对两个链表的连接和结尾处理。

代码

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public ListNode partition(ListNode head, int x) {        if(head == null){            return null;        }        ListNode dummy1 = new ListNode(Integer.MIN_VALUE);        ListNode dummy2 = new ListNode(Integer.MAX_VALUE);        ListNode cur1 = dummy1;        ListNode cur2 = dummy2;                ListNode cur = head;        while(cur!=null){            if(cur.val>=x){                cur2.next = cur;                cur2 = cur2.next;            }            else{                cur1.next = cur;                cur1 = cur1.next;            }            cur = cur.next;        }        cur1.next = dummy2.next;        cur2.next = null;        return dummy1.next;    }