分隔链表

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

 

示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5
通过次数58,794提交次数97,797

利用内存换时间和代码复杂度的方式

使用两个链表

1. 一个链表保存小于分区值的值，另一个链表保存大于分区值的值。

2. 小链表的尾和大链表的头结合

   /**
    * 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){
               return null;
           }
           ListNode leHead=new ListNode(0);
           ListNode gtHead=new ListNode(0);
           ListNode leFirst=leHead;
           ListNode gtFirst=gtHead;
           while(head!=null){
               if(head.val<x){
                   leHead.next=new ListNode(head.val);
                   leHead=leHead.next;
               }else {
                   gtHead.next=new ListNode(head.val);
                   gtHead=gtHead.next;
               }
               head=head.next;
           }
           leHead.next=gtFirst.next;
           return leFirst.next;
       }
   }

官方题解中，使用了双链表的思想，没有使用额外的辅组空间

使用两个哑节点，小于 $x$ 的放到第一个节点，大于 $x$ 的放到第二个节点

class Solution {
    public ListNode partition(ListNode head, int x) {

        // before and after are the two pointers used to create the two list
        // before_head and after_head are used to save the heads of the two lists.
        // All of these are initialized with the dummy nodes created.
        ListNode before_head = new ListNode(0);
        ListNode before = before_head;
        ListNode after_head = new ListNode(0);
        ListNode after = after_head;

        while (head != null) {

            // If the original list node is lesser than the given x,
            // assign it to the before list.
            // 这里没有释放节点的原因是下一次赋值时覆盖了原来的位置的元素
            if (head.val < x) {
                before.next = head;
                before = before.next;
                
            } else {
                // If the original list node is greater or equal to the given x,
                // assign it to the after list.
                after.next = head;
                after = after.next;
            }

            // move ahead in the original list
            head = head.next;
        }

        // Last node of "after" list would also be ending node of the reformed list
        // 这里必须要释放节点，否则就会和后面的节点链上
        after.next = null;

        // Once all the nodes are correctly assigned to the two lists,
        // combine them to form a single list which would be returned.
        before.next = after_head.next;

        return before_head.next;
    }
}

反思 : 对链表的理解还是不够清楚

参考文献

官方题解

赏

谢谢你请我吃糖果

微信