力扣面试150题--分隔链表

day 39

题目描述

思路

遍历链表，每一个点与值比较，比值小就继续，比值大就放到链表尾部即可

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        if(head==null){
            return null;
        }
        ListNode fakehead=new ListNode();
        fakehead.next=head;
        ListNode z=head;
        while(z.next!=null){
            z=z.next;
        }
        ListNode before=fakehead;
        ListNode y=head;
        ListNode end=z;
        while(y!=end){
            if(y.val>=x){
                z.next=y;
                before.next=y.next;
                y=y.next;
                z=z.next;
                z.next=null;
            }
            else{
                before=before.next;
                y=y.next;
            }
        }
        if(y.val>=x){
                z.next=y;
                before.next=y.next;
                y=y.next;
                z=z.next;
                z.next=null;
            }
        return fakehead.next;
    }
}