LeetCode-Partition List

Description：
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.

Example:

Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5

题意：给定一个链表和一个值$x$，要求将链表中节点值小于$x$的那些节点排列在链表的首部，并且保持在原链表中的相对顺序；

解法：定义三个指针变量

• p：指向排序后的链表的最后一个节点
• head：用于遍历原链表
• pre：指向head的前一个节点

我们遍历原链表，将所有节点值小于$x$的节点从原链表中删除，并插入到排序后的链表中；这里要注意的是不管是原链表还是排序后的链表，我们都是使用的同一个链表，并没有额外申请新的空间；

例如，对于链表​​head = 1->4->3->2->5->2, x = 3​​，其中vHead为虚拟头结点；

Java

/**
 * 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) {
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode vHead = new ListNode(0);
        vHead.next = head;
        ListNode pre = vHead;
        ListNode p = vHead;
        while (head != null) {
            if (head.val < x) {
                if (p.next == head) {
                    p = p.next;
                    pre = pre.next;
                    head = head.next;
                } else {
                    ListNode tempHead = head.next;
                    pre.next = head.next;
                    ListNode tempP = p.next;
                    p.next = head;
                    head.next = tempP;
                    p = head;
                    head = tempHead;
                }
            } else {
                pre = pre.next;
                head = head.next;
            }
        }
        
        return vHead.next;
    }
}