题目描述
给定单个链表的头 head
,使用 插入排序 对链表进行排序,并返回 排序后链表的头 。
插入排序 算法的步骤:
- 插入排序是迭代的,每次只移动一个元素,直到所有元素可以形成一个有序的输出列表。
- 每次迭代中,插入排序只从输入数据中移除一个待排序的元素,找到它在序列中适当的位置,并将其插入。
- 重复直到所有输入数据插入完为止。
下面是插入排序算法的一个图形示例。部分排序的列表(黑色)最初只包含列表中的第一个元素。每次迭代时,从输入数据中删除一个元素(红色),并就地插入已排序的列表中。
对链表进行插入排序。
示例 1:
输入: head = [4,2,1,3]
输出: [1,2,3,4]
示例 2:
输入: head = [-1,5,3,4,0]
输出: [-1,0,3,4,5]
提示:
- 列表中的节点数在
[1, 5000]
范围内
-5000 <= Node.val <= 5000
解法
方法一
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24 | # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def insertionSortList(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
dummy = ListNode(head.val, head)
pre, cur = dummy, head
while cur:
if pre.val <= cur.val:
pre, cur = cur, cur.next
continue
p = dummy
while p.next.val <= cur.val:
p = p.next
t = cur.next
cur.next = p.next
p.next = cur
pre.next = t
cur = t
return dummy.next
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36 | /**
* 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 insertionSortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(head.val, head);
ListNode pre = dummy, cur = head;
while (cur != null) {
if (pre.val <= cur.val) {
pre = cur;
cur = cur.next;
continue;
}
ListNode p = dummy;
while (p.next.val <= cur.val) {
p = p.next;
}
ListNode t = cur.next;
cur.next = p.next;
p.next = cur;
pre.next = t;
cur = t;
}
return dummy.next;
}
}
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34 | /**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var insertionSortList = function (head) {
if (head == null || head.next == null) return head;
let dummy = new ListNode(head.val, head);
let prev = dummy,
cur = head;
while (cur != null) {
if (prev.val <= cur.val) {
prev = cur;
cur = cur.next;
continue;
}
let p = dummy;
while (p.next.val <= cur.val) {
p = p.next;
}
let t = cur.next;
cur.next = p.next;
p.next = cur;
prev.next = t;
cur = t;
}
return dummy.next;
};
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