147. Insertion Sort List

题目：

Sort a linked list using insertion sort. 

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 Linked List Sort 

链接： https://leetcode.com/problems/insertion-sort-list/

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;
    }
}