insertion sort pseudocode

https://oj.leetcode.com/problems/insertion-sort-list/http://blog.csdn.net/linhuanmars/article/details/21144553/***definitionforsingly-linkedlist.*publicclasslistnode {*intval;*ListNode Next;*listnode (intx) {* val=x;*next =null;*}*}*/publicclasssolution {publiclistnodeinsertionsortlist (Listnodehead) {if (head==null| | head.next==null) returnhead;//adddummyhead listNodedummyhead=newlistnode (0); listnode

The insertion sort in the introduction to algorithms is handy. The core idea is quasi, the realization is prevail.#include 　　Algorithm does not return to the insertion sort (C version)

LeetCode -- Insertion Sort ListDescription:Sort a linked list using insertion sort.Ideas:Insert a sort list class, add the records to the list one by one, and use the list to create a new linked list.Implementation Code: /** * Definition for singly-linked list. * public class ListNode { * public int val; * publ

Sorts the linked list with an insert sort.See: https://leetcode.com/problems/insertion-sort-list/description//** * Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * listnode (int x): Val (x), Next (NULL) {} *}; */class Soluti On {public: listnode* insertionsortlist (listnode* head) { listnode* helper=new listnode ( -1); listnode* Cur=helper;

Package Algorithm;public class Straightinsertsort {/** * plus Sentinel Direct insertion sort * @param arr */public static void Insertsort (int[] ar R) {int i,j;for (i=2;i　　The direct insertion sort of the surveillance whistle

] is moved back one bit, at this point, the post-shifted bit covers the key, and then i--, and loop repeat, until the end of the loop, the end of the loop , from small to large arrangement, but at this time because of the previous i--reasons, after jumping out of the loop, the first step to save the key value to a[i+1]; must be i+1. This ensures that the values to be sorted are changed from left to right after this sort.The analysis of the text is still not straightforward enough to be understoo

} }return true; } Bstnode*Ptr= *Ptreeif(x==Ptr -Data) {return false; }Else if(xPtr -Data) {returnInsert (Ptr -Leftchild, x, PTR); }Else{returnInsert (Ptr -Rightchild, x, PTR); }}/** Iterative Implementation * /BOOL Insertbst (Bstree*Ptree,const Elemtype val) {if(Ptree== NULL)return false;if(*Ptree== NULL) {Bstnode*S=Buynode (); S -Data =Val S -Leftchild=S -Rightchild= NULL;*Ptree=S S -Parent = NULL;return true; } Bstnode*Ptr= *Ptree Bstnode*Pa=ptr while(PTR!= NULL) {if(PTR -Data

not change the seatif (Array[i] int sign = 0; Set the signal, default first, if larger than the previous element, then insert in the element seatfor (int j = i; j > 0; j--) {//Search from back to forwardif (Array[i] > Array[j-1]) {sign = j;Break}}Place the value of the signal at the TEMP variable, insert the value of array[i] after the signal elementtemp = array[sign];Array[sign] = Array[i];Count + = 1;Swaps the values so that the elements in front of the array I and behind sign are shifted one

Simple insert sort, always time out, leave this record for a moment.Class solution: # @param head, a listnode # @return a listnode def insertionsortlist (self, head): if head = = No NE or head.next = = None:return Headpsuhead = ListNode ( -1) while head:tail = Psuheadheadnext = Head.nextwhile Tail.next and Tail.next.val Last executed input: {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,2

[TOC] Inserting a Sort program codePackage Com.uplooking.bigdata.datastructure;import Java.util.arrays;public class Insertsort {public static void main (S Tring[] args) {int[] arr = {8,-2, 3, 9, 0, 1, 7, 6}; System.out.println ("Before sorting:" + arrays.tostring (arr)); Insertsort (arr); System.out.println ("After sorting:" + arrays.tostring (arr)); }/** * Insert sort: * View

Test instructions: Give a list of links to implement the insertion sort.Idea: O (1) space, O (n*n) complexity. String up the row with the other end of the list, and the linked list is deleted and returned to the list.1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7 * };8 */9 classSolution {Ten Public: Onelistnode* Insertionsortlist (listnode*head) { A

;Head = head->next;Nh->next = NULL;while (Head!=null) {if (Head->val ListNode *a = head->next;Head->next = NH;NH = head;head = A;}else {if (P->val > Head->val) {p = NH;}while (P->next head->val > P->next->val) {p = p->next;}if (P->next) {ListNode *a = head->next;Head->next = p->next;P->next = head;head = A;}else {P->next = head;Head = head->next;P->next->next = NULL;}}}return NH;}Efficiency comparison: Improved before the test speed of 80ms, improved test speed of 24ms, saving 70% of the time!

DescriptionSort a linked list using insertion sort.ExampleGiven 1->3->2->0->null , return 0->1->2->3->null .Problem solving: The list is sorted by inserting. A very simple topic, but there are still a lot of problems.Summarize the problems encountered:(1) Without a head node, in order to facilitate the construction of a, return to the first node of the next on the line.(2) There is no need to always in the original linked list of tangled, can apply fo

Problem description: Sort a linked list using insertion sort. Solution: 1 public class Solution { 2 public ListNode insertionSortList(ListNode head) { 3 int length = 0; 4 ListNode p = head; 5 while (p != null) { 6 length++; 7 p = p.next; 8 } 9 10 if (length

Error-prone points1 not in situ sort, surprised!!2 to reposition pre!! Public classSolution { PublicListNode insertionsortlist (ListNode head) {//It 's easy to figure out how to insert it. if(head==NULL|| Head.next = =NULL)returnHead; ListNode h=NewListNode (-1); //h.next = head;//not in-situ sortingListNode cur =Head; while(cur!=NULL){ //to redefine the point of the Pre!!! ListNode pre =h; ListNode T=Cur.next; while(pre.next!=NULL

The method of directly inserting the sort (straight insertion sort) is: Each time the first element is extracted from the unordered table, it is inserted to the proper position of the ordered table, so that the ordered table is still ordered. The procedure is as follows: Initial sequence:I = 1 [46] 58 15 45 90 18 10 62BytesI = 2 [46 58] 15 45 90 18 10 62Sna

Sort a linked list using insertion sort. Class solution {public: listnode * insertionsortlist (listnode * head) {If (Head = NULL | head-> next = NULL) return head; listnode * result; result-> val = int_min; Result-> next = NULL; listnode * cur = head, * POs, * pre; while (cur! = NULL) {pos = Result-> next; Pre = result; while (Pos! = NULL Pos-> Val

Package Net.qh.test.sort;import java.util.arraylist;import Java.util.calendar;import java.util.list;/** * Created by Administrator on 2016/03/01. */public class Insert {public int[] sort (int[] arr) {if (arr = = NULL | | arr.length 　　A sort algorithm of Java insertion sorting algorithm

Interview 24: Encode direct Insert sort the original answer:#include Thought analysis: The whole idea no matter what mistake. Error analysis such as the following: for (i = 2; I array[0]//selected as Sentry causes the first number in the sequence to always change and cannot be sorted correctly >>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>>>>>>>>>>>>>>>>>>>>>>> >>>>>> Fix the answer: #incl