题目：

Sort a linked list using insertion sort. 

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题解：

链表插入排序，考察基本功。worst case时间复杂度是O(n²) ， best case时间复杂度是O(n)。   好久没做题最近完全懈怠了，要继续努力啊。

Time Complexity - O(n²) (worst case)， Space Complexity - O(n)。

public class Solution {    public ListNode insertionSortList(ListNode head) {        if(head == null || head.next == null)            return head;        ListNode dummy = new ListNode(-1);                while(head != null){            ListNode node = dummy;                        while(node.next != null && node.next.val <= head.val)                node = node.next;                        ListNode temp = head.next;            head.next = node.next;            node.next = head;            head = temp;        }                return dummy.next;    }}

 

Update:

/** * 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) {        if(head == null || head.next == null)            return head;        ListNode dummy = new ListNode(-1);                while(head != null) {            ListNode node = dummy;                        while(node.next != null && node.next.val < head.val )                 node = node.next;                        ListNode temp = head.next;            head.next = node.next;            node.next = head;            head = temp;        }                return dummy.next;    }}

 

二刷:

跟一刷使用的节点交换顺序不太一样。不过大体方法差不多。在head不为空的情况下，先设一个dummy的reference copy - node，从头比较node.next.val 和head.val来找到insert的位置，再倒腾几下就好了。

Java:

Time Complexity - O(n²) (worst case)， Space Complexity - O(n)。

/** * 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(-1);        while (head != null) {            ListNode node = dummy;            while (node.next != null && node.next.val < head.val) {                node = node.next;            }            ListNode tmp = node.next;            node.next = head;             head = head.next;            node.next.next = tmp;        }        return dummy.next;    }}

 

三刷:

Java:

Time Complexity - O(n²) (worst case)， Space Complexity - O(1)。

/** * 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(-1);        ListNode tmp = null, node = dummy;;        while (head != null) {            node = dummy;            while (node.next != null && node.next.val < head.val) node = node.next;            tmp = node.next;            node.next = head;            head = head.next;            node = node.next;            node.next = tmp;        }        return dummy.next;    }}