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Implementation of various sorting algorithms in Java

1. Direct insert sort

This sort of sorting problem is often encountered : Insert the new data into the already arranged data column .

  1. Sort the first number and the second number , And then form an ordered sequence
  2. Insert the third number in , Form a new ordered sequence .
  3. For the fourth number 、 The fifth number …… Until the last number , Repeat step 2 . 

How to write code :

  1. First set the number of inserts , The number of cycles ,for(int i=1;i<length;i++),1 No need to insert the number .
  2. Set the number of inserts and the number of digits to get the last number of the ordered sequence .insertNum and j=i-1.
  3. Cycle forward from the last number , If the number of inserts is less than the current number , Just move the current number back one bit .
  4. Put the current number in an empty position , namely j+1.

The code implementation is as follows :

public void insertSort(int[] a){
        int length=a.length;// The length of the array , This is extracted to speed up .
        int insertNum;// The number to insert 
        for(int i=1;i<length;i++){// Number of inserts 
            insertNum=a[i];// The number to insert 
            int j=i-1;// The number of ordered sequence elements 
            while(j>=0&&a[j]>insertNum){// The sequence circulates from back to front , Will be bigger than the insertNum Move the number back one space 
                a[j+1]=a[j];// Element moves one space 
                j--;
            }
            a[j+1]=insertNum;// Put the number to be inserted in the position to be inserted .
        }
    }

2. Shell Sort

For the direct insertion sort problem , When the amount of data is huge .

  1. Set the number of numbers to n, Take an odd number k=n/2, The difference between subscripts is k The books are divided into a group , Constitute an ordered sequence .
  2. Retake k=k/2 , The difference between subscripts is k The books are divided into a group , Constitute an ordered sequence .
  3. Repeat step 2 , until k=1 Perform simple insert sort .

 

How to write code :

  1. First determine the number of components .
  2. Then insert and sort the elements in the group .
  3. And then length/2, repeat 1,2 Step , until length=0 until .

The code implementation is as follows :

public  void sheelSort(int[] a){
        int d  = a.length;
        while (d!=0) {
            d=d/2;
            for (int x = 0; x < d; x++) {// Number of components 
                for (int i = x + d; i < a.length; i += d) {// Elements in groups , Start with the second number 
                    int j = i - d;//j The last digit of an ordered sequence 
                    int temp = a[i];// Elements to insert 
                    for (; j >= 0 && temp < a[j]; j -= d) {// Go back and forth .
                        a[j + d] = a[j];// Move backward d position 
                    }
                    a[j + d] = temp;
                }
            }
        }
    }

3. Simple selection sort

It is often used to take the maximum and minimum number in a sequence .

( If every comparison is exchanged , So it's exchange ordering ; If you compare one cycle at a time and then exchange , It's a simple selection and sorting .)

  1. Traverse the entire sequence , Put the minimum number first .
  2. Traverse the rest of the sequence , Put the minimum number first .
  3. Repeat step 2 , Until there is only one number left .

 

How to write code :

  1. First determine the number of cycles , And remember the current number and position .
  2. Compare all the numbers after the current position with the current number , The decimal is assigned to key, And remember the decimal place .
  3. After the comparison , Exchange the minimum value with the value of the first number .
  4. repeat 2、3 Step .

The code implementation is as follows :

public void selectSort(int[] a) {
        int length = a.length;
        for (int i = 0; i < length; i++) {// cycles 
            int key = a[i];
            int position=i;
            for (int j = i + 1; j < length; j++) {// Select the minimum value and position 
                if (a[j] < key) {
                    key = a[j];
                    position = j;
                }
            }
            a[position]=a[i];// Swap places 
            a[i]=key;
        }
    }

 

4. Heap sort

Optimization of simple selection and sorting .

  1. Build the sequence into a big top heap .
  2. Exchange the root node with the last node , Then disconnect the last node .
  3. Repeat first 、 Two steps , Until all nodes are disconnected .

 

The code implementation is as follows :

public  void heapSort(int[] a){
        System.out.println(" Start sorting ");
        int arrayLength=a.length;
        // Cycle building   
        for(int i=0;i<arrayLength-1;i++){
            // Building the heap   

            buildMaxHeap(a,arrayLength-1-i);
            // Swap top and last elements   
            swap(a,0,arrayLength-1-i);
            System.out.println(Arrays.toString(a));
        }
    }
    private  void swap(int[] data, int i, int j) {
        // TODO Auto-generated method stub  
        int tmp=data[i];
        data[i]=data[j];
        data[j]=tmp;
    }
    // Yes data Array from 0 To lastIndex Build a large top pile   
    private void buildMaxHeap(int[] data, int lastIndex) {
        // TODO Auto-generated method stub  
        // from lastIndex Nodes ( The last node ) Start of parent node of   
        for(int i=(lastIndex-1)/2;i>=0;i--){
            //k Save the node being judged   
            int k=i;
            // If at present k Child node of node exists   
            while(k*2+1<=lastIndex){
                //k Index of the left child of the node   
                int biggerIndex=2*k+1;
                // If biggerIndex Less than lastIndex, namely biggerIndex+1 Representative k The right child of the node exists   
                if(biggerIndex<lastIndex){
                    // If the value of the right child node is large   
                    if(data[biggerIndex]<data[biggerIndex+1]){
                        //biggerIndex Always record the index of the larger child node   
                        biggerIndex++;
                    }
                }
                // If k The value of a node is less than the value of its larger child nodes   
                if(data[k]<data[biggerIndex]){
                    // Exchange them   
                    swap(data,k,biggerIndex);
                    // take biggerIndex give k, Start while Next cycle of the cycle , Reassurance k The value of a node is greater than the value of its left and right child nodes   
                    k=biggerIndex;
                }else{
                    break;
                }
            }
        }
    }

5. Bubble sort

Generally do not use .

  1. Compare all elements of the sequence in pairs , Put the biggest one at the back .
  2. Compare all elements of the remaining sequence in pairs , Put the biggest one at the back .
  3. Repeat step 2 , Until there is only one number left .

 

How to write code :

  1. Set the number of cycles .
  2. Set the number of digits to start the comparison , And the number of digits at the end .
  3. Comparing the two , Put the smallest in front .
  4. repeat 2、3 Step , Until the number of cycles is over .

The code implementation is as follows :

public void bubbleSort(int[] a){
        int length=a.length;
        int temp;
        for(int i=0;i<a.length;i++){
            for(int j=0;j<a.length-i-1;j++){
                if(a[j]>a[j+1]){
                    temp=a[j];
                    a[j]=a[j+1];
                    a[j+1]=temp;
                }
            }
        }
    }

6. Quick sort

Ask for the fastest time .

  1. Choose the first number as p, Less than p The number of is on the left , Greater than p The number of is on the right .
  2. Recursively will p The numbers on the left and right follow the first step , Until you can't recurse .

 

The code implementation is as follows :

public static void quickSort(int[] numbers, int start, int end) {   
    if (start < end) {   
        int base = numbers[start]; //  Selected reference value ( The first value is the reference value )   
        int temp; //  Record temporary intermediate value    
        int i = start, j = end;   
        do {   
            while ((numbers[i] < base) && (i < end))   
                i++;   
            while ((numbers[j] > base) && (j > start))   
                j--;   
            if (i <= j) {   
                temp = numbers[i];   
                numbers[i] = numbers[j];   
                numbers[j] = temp;   
                i++;   
                j--;   
            }   
        } while (i <= j);   
        if (start < j)   
            quickSort(numbers, start, j);   
        if (end > i)   
            quickSort(numbers, i, end);   
    }   
}

7. Merge sort

Second only to fast platoon , Use when memory is low , Can be used when parallel computing .

  1. Choose two adjacent numbers to form an ordered sequence .
  2. Select two adjacent ordered sequences to form an ordered sequence .
  3. Repeat step 2 , Until it all forms an ordered sequence . 

  The code implementation is as follows :

public static void mergeSort(int[] numbers, int left, int right) {   
    int t = 1;//  Number of elements in each group    
    int size = right - left + 1;   
    while (t < size) {   
        int s = t;//  The number of elements in each group of this cycle    
        t = 2 * s;   
        int i = left;   
        while (i + (t - 1) < size) {   
            merge(numbers, i, i + (s - 1), i + (t - 1));   
            i += t;   
        }   
        if (i + (s - 1) < right)   
            merge(numbers, i, i + (s - 1), right);   
    }   
}   
private static void merge(int[] data, int p, int q, int r) {   
    int[] B = new int[data.length];   
    int s = p;   
    int t = q + 1;   
    int k = p;   
    while (s <= q && t <= r) {   
        if (data[s] <= data[t]) {   
            B[k] = data[s];   
            s++;   
        } else {   
            B[k] = data[t];   
            t++;   
        }   
        k++;   
    }   
    if (s == q + 1)   
        B[k++] = data[t++];   
    else  
        B[k++] = data[s++];   
    for (int i = p; i <= r; i++)   
        data[i] = B[i];   
}

8. Radix sorting

For a large number of , When a long number is sorted .

  1. Take out the single digits of all numbers , Sort by single digits , Make a sequence .
  2. Take out the tens of all the new numbers , Sort by tens , Make a sequence .

The code implementation is as follows :

public void sort(int[] array) {
        // First, determine the number of times to sort ;     
        int max = array[0];
        for (int i = 1; i < array.length; i++) {
            if (array[i] > max) {
                max = array[i];
            }
        }
        int time = 0;
        // Determine the number of digits ;     
        while (max > 0) {
            max /= 10;
            time++;
        }
        // establish 10 A queue ;     
        List<ArrayList> queue = new ArrayList<ArrayList>();
        for (int i = 0; i < 10; i++) {
            ArrayList<Integer> queue1 = new ArrayList<Integer>();
            queue.add(queue1);
        }
        // Conduct time Sub distribution and collection ;     
        for (int i = 0; i < time; i++) {
            // Assign array elements ;     
            for (int j = 0; j < array.length; j++) {
                // Get the number time+1 digit ;   
                int x = array[j] % (int) Math.pow(10, i + 1) / (int) Math.pow(10, i);
                ArrayList<Integer> queue2 = queue.get(x);
                queue2.add(array[j]);
                queue.set(x, queue2);
            }
            int count = 0;// Element counter ;     
            // Collect queue elements ;     
            for (int k = 0; k < 10; k++) {
                while (queue.get(k).size() > 0) {
                    ArrayList<Integer> queue3 = queue.get(k);
                    array[count] = queue3.get(0);
                    queue3.remove(0);
                    count++;
                }
            }
        }
    }


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