147. 对链表进行插入排序

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

Python3JavaJavaScriptGo

# 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

/**
 * 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;
    }
}

/**
 * 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;
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func insertionSortList(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    dummy := &ListNode{head.Val, head}
    pre, cur := dummy, head
    for cur != nil {
        if pre.Val <= cur.Val {
            pre = cur
            cur = cur.Next
            continue
        }
        p := dummy
        for 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
}

147. 对链表进行插入排序

题目描述

解法

方法一

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