LeetCode.86 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.

分析：

class Solution {    public ListNode partition(ListNode head, int x) {        //根据给定值划分链区，小的在左边，大的在其后面(小的放前面，大的放后面)//小的插入头结点后面，大于等于不移动        ListNode dummy = new ListNode(0);dummy.next = head;ListNode res = dummy;ListNode pre = dummy;ListNode cur = head;while (cur != null) {if (cur.val < x) {// 小于pre.next = cur.next;ListNode temp = pre.next;cur.next = dummy.next;dummy.next = cur;dummy = dummy.next;if(pre.next==cur){//判断当前节点是否为原位置pre=pre.next;}cur = temp;} else {// 大于等于pre = pre.next;cur = cur.next;}}return res.next;    }}

参考答案：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode partition(ListNode head, int x) {        //利用两个链表，一个用来装小于的，另一个用来装大于或者等于。        //最后通过拼接两个链表，返回结果        ListNode dummy1 = new ListNode(0);//小的        ListNode dummy2 = new ListNode(0); //大于等于         ListNode curr1 = dummy1, curr2 = dummy2; //标志位          while (head!=null){            if (head.val<x) {                curr1.next = head;                curr1 = head;            }else {                curr2.next = head;                curr2 = head;            }            head = head.next;        }        //拼接,curr2.next置空防止循环        curr2.next = null;                 curr1.next = dummy2.next;        return dummy1.next;    }}