🔥码云GVP开源项目 12k star Uniapp+ElementUI 功能强大 支持多语言、二开方便! 广告
# Insertion Sort List ### Source - leetcode: [Insertion Sort List | LeetCode OJ](https://leetcode.com/problems/insertion-sort-list/) - lintcode: [(173) Insertion Sort List](http://www.lintcode.com/en/problem/insertion-sort-list/) ~~~ Sort a linked list using insertion sort. Example Given 1->3->2->0->null, return 0->1->2->3->null. ~~~ ### 题解1 - 从首到尾遍历 插入排序常见的实现是针对数组的,如前几章总的的 [Insertion Sort](http://algorithm.yuanbin.me/zh-cn/basics_sorting/insertion_sort.html),但这道题中的排序的数据结构为单向链表,故无法再从后往前遍历比较值的大小了。好在天无绝人之路,我们还可以**从前往后依次遍历比较和交换。** 由于排序后头节点不一定,故需要引入 dummy 大法,并以此节点的`next`作为最后返回结果的头节点,返回的链表从`dummy->next`这里开始构建。首先我们每次都从`dummy->next`开始遍历,依次和上一轮处理到的节点的值进行比较,直至找到不小于上一轮节点值的节点为止,随后将上一轮节点插入到当前遍历的节点之前,依此类推。文字描述起来可能比较模糊,大家可以结合以下的代码在纸上分析下。 ### Python ~~~ """ Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: The first node of linked list. @return: The head of linked list. """ def insertionSortList(self, head): dummy = ListNode(0) cur = head while cur is not None: pre = dummy while pre.next is not None and pre.next.val < cur.val: pre = pre.next temp = cur.next cur.next = pre.next pre.next = cur cur = temp return dummy.next ~~~ ### C++ ~~~ /** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @return: The head of linked list. */ ListNode *insertionSortList(ListNode *head) { ListNode *dummy = new ListNode(0); ListNode *cur = head; while (cur != NULL) { ListNode *pre = dummy; while (pre->next != NULL && pre->next->val < cur->val) { pre = pre->next; } ListNode *temp = cur->next; cur->next = pre->next; pre->next = cur; cur = temp; } return dummy->next; } }; ~~~ ### Java ~~~ /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode insertionSortList(ListNode head) { ListNode dummy = new ListNode(0); ListNode cur = head; while (cur != null) { ListNode pre = dummy; while (pre.next != null && pre.next.val < cur.val) { pre = pre.next; } ListNode temp = cur.next; cur.next = pre.next; pre.next = cur; cur = temp; } return dummy.next; } } ~~~ ### 源码分析 1. 新建 dummy 节点,用以处理最终返回结果中头节点不定的情况。 1. 以`cur`表示当前正在处理的节点,在从 dummy 开始遍历前保存`cur`的下一个节点作为下一轮的`cur`. 1. 以`pre`作为遍历节点,直到找到不小于`cur`值的节点为止。 1. 将`pre`的下一个节点`pre->next`链接到`cur->next`上,`cur`链接到`pre->next`, 最后将`cur`指向下一个节点。 1. 返回`dummy->next`最为最终头节点。 Python 的实现在 lintcode 上会提示 [TLE](# "Time Limit Exceeded 的简称。你的程序在 OJ 上的运行时间太长了,超过了对应题目的时间限制。"), leetcode 上勉强通过,这里需要注意的是采用`if A is not None:`的效率要比`if A:`高,不然 leetcode 上也过不了。具体原因可参考 [Stack Overflow](http://stackoverflow.com/questions/7816363/if-a-vs-if-a-is-not-none) 上的讨论。 ### 复杂度分析 最好情况:原链表已经有序,每得到一个新节点都需要 iii 次比较和一次交换, 时间复杂度为 1/2O(n2)+O(n)1/2O(n^2) + O(n)1/2O(n2)+O(n), 使用了 dummy 和 pre, 空间复杂度近似为 O(1)O(1)O(1). 最坏情况:原链表正好逆序,由于是单向链表只能从前往后依次遍历,交换和比较次数均为 1/2O(n2)1/2 O(n^2)1/2O(n2), 总的时间复杂度近似为 O(n2)O(n^2)O(n2), 空间复杂度同上,近似为 O(1)O(1)O(1). ### 题解2 - 优化有序链表 从题解1的复杂度分析可以看出其在最好情况下时间复杂度都为 O(n2)O(n^2)O(n2) ,这显然是需要优化的。 仔细观察可发现最好情况下的比较次数 是可以优化到 O(n)O(n)O(n) 的。思路自然就是先判断链表是否有序,仅对降序的部分进行处理。优化之后的代码就没题解1那么容易写对了,建议画个图自行纸上分析下。 ### Python ~~~ """ Definition of ListNode class ListNode(object): def __init__(self, val, next=None): self.val = val self.next = next """ class Solution: """ @param head: The first node of linked list. @return: The head of linked list. """ def insertionSortList(self, head): dummy = ListNode(0) dummy.next = head cur = head while cur is not None: if cur.next is not None and cur.next.val < cur.val: # find insert position for smaller(cur->next) pre = dummy while pre.next is not None and pre.next.val < cur.next.val: pre = pre.next # insert cur->next after pre temp = pre.next pre.next = cur.next cur.next = cur.next.next pre.next.next = temp else: cur = cur.next return dummy.next ~~~ ### C++ ~~~ /** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @return: The head of linked list. */ ListNode *insertionSortList(ListNode *head) { ListNode *dummy = new ListNode(0); dummy->next = head; ListNode *cur = head; while (cur != NULL) { if (cur->next != NULL && cur->next->val < cur->val) { ListNode *pre = dummy; // find insert position for smaller(cur->next) while (pre->next != NULL && pre->next->val <= cur->next->val) { pre = pre->next; } // insert cur->next after pre ListNode *temp = pre->next; pre->next = cur->next; cur->next = cur->next->next; pre->next->next = temp; } else { cur = cur->next; } } return dummy->next; } }; ~~~ ### Java ~~~ /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode insertionSortList(ListNode head) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode cur = head; while (cur != null) { if (cur.next != null && cur.next.val < cur.val) { // find insert position for smaller(cur->next) ListNode pre = dummy; while (pre.next != null && pre.next.val < cur.next.val) { pre = pre.next; } // insert cur->next after pre ListNode temp = pre.next; pre.next = cur.next; cur.next = cur.next.next; pre.next.next = temp; } else { cur = cur.next; } } return dummy.next; } } ~~~ ### 源码分析 1. 新建 dummy 节点并将其`next` 指向`head` 1. 分情况讨论,仅需要处理逆序部分。 1. 由于已经确认链表逆序,故仅需将较小值(`cur->next`而不是`cur`)的节点插入到链表的合适位置。 1. 将`cur->next`插入到`pre`之后,这里需要四个步骤,需要特别小心! ![Insertion Sort](https://box.kancloud.cn/2015-10-24_562b1f502445b.png) 如上图所示,将`cur->next`插入到`pre`节点后大致分为3个步骤。 ### 复杂度分析 最好情况下时间复杂度降至 O(n)O(n)O(n), 其他同题解1. ### Reference - [Explained C++ solution (24ms) - Leetcode Discuss](https://leetcode.com/discuss/37574/explained-c-solution-24ms) - [Insertion Sort List - 九章算法](http://www.jiuzhang.com/solutions/insertion-sort-list/)