leetcode——86. 分隔链表

慢慢找到对链表的感觉，加油保持前进呀哈哈哈哈！！！！

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def partition(self, head, x):
        """
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        """
        if not head:
            return 
        a=ListNode(-1)
        a.next=head
        p=head
        r=p
        q=p.next
        if p.val>=x:
            p=a
        while q:
            if q.val>=x:
                q=q.next
                r=r.next
            else:
                if q==p.next:
                    p=p.next
                    r=r.next
                    q=q.next
                else:
                    r.next=q.next
                    q.next=p.next
                    p.next=q
                    p=q
                    q=r.next
        return a.next

执行用时 :20 ms, 在所有 python 提交中击败了91.08%的用户

内存消耗 :11.8 MB, 在所有 python 提交中击败了26.74%的用户

 

——2019.10.26

 

---

public ListNode partition(ListNode head, int x) {
        if(head == null || head.next == null){
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode p = dummy;
        ListNode s = head;
        ListNode t = dummy;
        while (s != null && s.val < x){
            s = s.next;
            t = t.next;
            p = p.next;
        }
        while (s != null){
            if(s.val >= x){
                s = s.next;
                t = t.next;
            }else{
                t.next = s.next;
                s.next = p.next;
                p.next = s;
                p = p.next;
                s = t.next;
            }
        }
        return dummy.next;
    }

比较简单

 ——2020.7.14