My Java Development Learning journey------Binary Insertion sort of >java classic sorting algorithm

First, binary Insert Sort ( Two-point insertion sort )

the method of inserting the insertion position in the direct insertion sort to find a[i] is replaced by the binary comparison, and the binary insertion sorting algorithm can be obtained. When dealing with a[i], a[0] ... A[I-1] has been ordered by key code values. The so-called binary comparison, is when inserting a[i], take A[I-1/2] key code value and A[i] key code value, if A[I] key code value is less than A[I-1/2] key code value, then a[i] can only insert a[0] to A[I-1/2] between, it can be in a[0] Continue to use binary comparisons between a[i-1/2-1], otherwise you can only insert A[I-1/2] to a[i-1], so a[i-1/2+1 comparisons may continue between a[i-1] to binary. This will be done until the position of insertion is determined at the end. Generally in a[k] and a[r] between the use of binary, wherein the node is A[K+R/2], after a comparison can be ruled out half of the record, the possible insertion of the interval reduced by half, so called binary. The prerequisite for performing a binary insert sort is that the file records must be stored sequentially.

Second, the principle of the algorithm

binary algorithm idea for inserting sorting:the basic process of the algorithm:(1) Calculate the middle point of 0 ~ i-1, with the element at the I index and the median value, if the element at the I index is large, indicating that the element to be inserted between the middle value and just added I index, and vice versa, is at the beginning of the position to the median position, so it is very simple to complete the binary;(2) in the corresponding half of the area to find the insertion position, the continuous use of (1) steps to narrow the range, non-stop binary, the range is reduced to 1/2 1/4 1/8 .... Quickly determine where the first element is to be inserted;(3) After the position is determined, the entire sequence is moved back and the element is inserted into the appropriate position.



Third, the code implementation

public class Binarysort {public static void Binarysort (int[] source) {int I, j;int high, Low, mid;int temp;for (i = 1; I & Lt Source.length; i++) {//Lookup area upper Bound low = 0;//Lookup area lower bound high = i-1;//Save the currently pending record in temporary variable temp = source[i];while (Low <= High) {//Find middle value/mid = (lo W + high)/2;mid = (low + high) >> 1;//if the record to be inserted is less than the intermediate record if (Temp<source[mid]) {//the insertion point is in the lower half of the zone high = mid-1;} else {// Insertion point in high half area low = mid + 1;}} Move all previous records that are larger than the current record to be inserted for (j = i-1; J >=low; j--) {source[j + 1] = source[j];} Backfill the record to be inserted in the correct position. Source[low] = temp; System.out.print ("First" + i + "sequencing:");p Rintarray (source);}} private static void PrintArray (int[] source) {for (int i = 0; i < source.length; i++) {System.out.print ("\ t" + source[i ]);} System.out.println ();} public static void Main (string[] args) {int source[] = new int[] {12, 15, 9, 14, 4, 18, 23, 6}; System.out.print ("Initial keyword:");p Rintarray (source); System.out.println (""); Binarysort (source); System.out.print ("\ n-sort result:");p Rintarray (source);}}

Iv. Results of Operation:

Initial keyword: 1215914418236 the first 1 orders: 1215914418236 The first 2 order: 9121514418236 the 3 order: 9121415418236 the 4 order: 4912141518236 The 5 order: 4912141518236 The first 6 orders: 4 912141518236 Part 7 order: 4691214151823 sorted Results: 4691214151823

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My Java Development Learning journey------Binary Insertion sort of >java classic sorting algorithm