Java implementation of binary (binary) Insert sort

With a sequence a[0],a[1]...a[n], where a[i-1] is already ordered, when inserting a[i], the binary search a[i] Insert position

Efficiency: O (n^2), for the initial basic ordered sequence, the efficiency is inferior to the direct insertion sort; for random unordered sequences, the efficiency is higher than the direct insertion sort.

/* * BINARY (binary) Insert sort * with a sequence a[0],a[1]...a[n], where a[i-1] is already ordered, when inserting a[i], use dichotomy to search A[i] Insert position */public class Binaryinsertsort { public static void Main (string[] args) {int len = 10;int[] ary = new Int[len]; Random random = new random (), for (int j = 0; J < Len; J + +) {Ary[j] = random.nextint (1000);} Binaryinsert (ary);/* Complexity analysis: the best case, that is, the order, there is no need to move right, the time complexity is: O (n lg N) The worst case, all in reverse order, when the complexity of O (n^2) * Can not increase the complexity of the worst case to O (N|logn). *///Print array printarray (ary);} /** * Insert Sort * @param ary */private static void Binaryinsert (int[] ary) {int setvaluecount = 0;//sort from the second element of the array, because the first element itself must have been Ordered for (int j = 1; j < Ary.length; J + +) {//complexity n//Save current value int key = ary[j];//? Use binary lookup to position the insertion position//int index = BinarySearch ASC (ary, Ary[j], 0, j-1);//complexity: O (logn)//int index = Binarysearchdesc (ary, Ary[j], 0, j-1);//complexity: O (logn) int index = b INARYSEARCHDESC2 (ary, Ary[j], 0, j-1);//Complexity: O (LOGN) printArray (ary); System.out.println (the position of the element on the "+ j +" Index to be inserted is: "+ index);//The target is inserted in the position, and the right side of the target position is moved to the right of the element for (int i = j; i > Index; i--) {//complexity, Worst Case: (n1) + (n-2) +...+n/2=o (n^2) ary[i] = ary[i-1]; I-1 <==> indexsetvaluecount++;} Ary[index] = key;setvaluecount++;} System.out.println ("\ n Set the number of values (Setvaluecount) =====>" + Setvaluecount);} /** * Binary Search Ascending recursion * * @param ary * Given sorted unknown origin array * @param target * Find target * @param from * current find The range starting point * @param to * The return end point of the current lookup * @return returns the destination in the array, in order where it should be */private static int binarysearchasc (int[] ary, int t arget, int from, int to) {int range = to-from;//If the range is greater than 0, that is, there are more than two elements, continue to split if (range > 0) {//select intermediate bit int mid = (to + from )/2;//If the critical bit is not met, then continue to find if (Ary[mid] > target) {/* * mid > Target, ascending rule, target smaller, should be swapped position in front, that is, target is positioned in the mid position, * According to the check Find ideas, from the from to the mid-1 think orderly, so to=mid-1 */return binarysearchasc (ary, Target, from, mid-1);} else {/* * Mid < Target, ascending rule, target larger, no swap position, lookup comparison starting position should be mid+1 */return binarysearchasc (ary, Target, mid + 1, to);}} else {if (Ary[from] > target) {//as 5,4, to insert 4return from;} else {return from + 1;}}} /** * Binary search Descending, recursive */private static int Binarysearchdesc (int[] ary, int target, int from, int to) {int range = to-from;if (Range > 0) {int mid = (from + to) >>> 1;if (Ary[mid] > target) {return Binarysearchdesc (ary, Target, mid + 1, to);} else {return bin Arysearchdesc (ary, Target, from, mid-1);}} else {if (Ary[from] > target) {//As 5,4, 4return from + 1 is to be inserted;} else {return from;}}} /** * Binary Lookup Descending, non-recursive */private static int binarySearchDesc2 (int[] ary, int target, int from, int. to) {//while (from < to) {for (; from < to;) {int mid = (from + to) >>> 1;if (Ary[mid] > target) {from = Mid + 1,} else {to = mid-1;}} From <==> to;if (Ary[from] > target) {//5,4, to insert 4return from + 1;} else {return from;}} private static void PrintArray (int[] ary) {for (int i:ary) {System.out.print (i + "");}}}

Print

Java implementation of binary (binary) Insert sort