Data structures and algorithms --7 search algorithm

1. Common Search Algorithm

1 ) order (linear) Find
 2 ) binary search / binary search
 3 ) Find interpolation
 4 ) Find Fibonacci

2. Find

1) linear search

A. Title:

There are a number of columns [ 1 , 43 , 22 - 10 , 0 ], the number of columns is determined whether to include this name, if found, to find the tips, and the index value is given.

B, ideas:

Find one by one

C. Code

com.offcn.search Package; 
// linear search 
public  class SeqSearch {
     public  static  void main (String [] args) {
         int [] = {ARR . 1 , 43 is , 22 is , - 10 , 0 };
         int Result = seqSearch (ARR , 0 );
         IF ! (Result = - . 1 ) { 
            . the System OUT .println ( " ! found subscript " + Result); 
        } the else { 
            . the System OUT .println ("未找到");
        }
    }
    public static int seqSearch(int[] arr,int value){
        for(int i = 0;i < arr.length;i++){
            if(arr[i] == value){
                return i;
            }
        }
        return -1;
    }
}

D. Output:

 

 2) binary search

A. Topic 1:

Please ordered array on a binary search { 1 , 8 , 10 , 89 , 1000 , 1234 }, enter a number to see whether this number of the array, and the subscript determined, if not prompt "this number is not"

B. ideas:

 C. Code

package com.offcn.search;
//二分查找
public class BinarySearch {
    public static void main(String[] args){
        int[] arr = {1,8,10,89,1000,1234};
        int i = binarySearch(arr, 0, arr.length - 1, 1234);
        if(i == -1){
            System.out.println("未找到");
        }the else { 
            the System. OUT .println ( " found subscript! " + I); 
        } 
    } 
    public  static  int binarySearch ( int [] ARR, int left, int right, int findValue) {
         // If not find, left It may have been mid + 1, out of range 
        IF (left> right) {
             return - . 1 ; 
        } 
        // intermediate array subscript 
        int mID = (left + right) / 2 ;
         int midValue =ARR [MID];
         // If the element is larger than the intermediate value to find, the recursive right 
        IF (findValue> midValue) {
             return binarySearch (ARR, MID + . 1 , right, findValue);
             // less than the intermediate value, left recursive 
        } the else  IF (findValue < midValue) {
             return binarySearch (ARR, left, mid- . 1 , findValue);
             // exactly intermediate value 
        } the else {
             return mID; 
        } 
    } 
}

D. Output:

 

 A. Question 2:

{ 1 , 8 , 10 , 89 , 1000 , 1000 , 1000 , 1000 , 1234 } When an ordered array of a plurality of the same value, all values are how to find?

B. ideas:

 

 C. Code

package com.offcn.search;

import java.util.ArrayList;

public class BinarySearchList {
    public static void main(String[] args){
        int[] arr = {1,8,10,89,1000,1000,1000,1000,1234};
        ArrayList<Integer> list = binarySearchList(arr, 0, arr.length - 1, 1000);
        System.out.println(list);

    }
    public static ArrayList<Integer> binarySearchList(int[] arr,int left,int right,int findValue){
        if(left > right){
            return new ArrayList<>();
        }

        int mid = (left+right)/2;
        int midValue= arr[mid];
        if(findValue > midValue){
            return binarySearchList(arr,mid+1,right,findValue);
        }else if(findValue < midValue){
            return binarySearchList (ARR, left, mid- . 1 , findValue); 
        } the else {
             // if found, does not return to, a collection means to create the same numerical subscript 
            the ArrayList List = new new the ArrayList ();
             // left side of the mid scan, to find the same value of the subscript added to the collection 
            int TEMP = mid- . 1 ;
             the while ( to true ) {
                 IF (TEMP < 0 ! || ARR [TEMP] = findValue) {
                     BREAK ; 
                } 
                List.add (TEMP); 
                the TEMP - = 1 ;
            } 
            List.add (mid); 
            // to the right of the mid scan, to find the same value of the subscript added to the collection 
            TEMP = mid + . 1 ;
             the while ( to true ) {
                 IF (TEMP> arr.length- . 1 || ARR [TEMP !] = findValue) {
                     BREAK ; 
                } 
                List.add (TEMP); 
                TEMP + = . 1 ; 
            } 
            return List; 
        } 
    } 
}

D. Output:

 

 3) Find interpolation

A. Principle:

B. Note:

C. Code

package com.offcn.search;
//插值查找
public class InsertValSearch {
    public static void main(String[] args){
        int[] arr = new int[100];
        for(int i = 0; i < arr.length;i++){
            arr[i] = i+1;
        }
        int i = insertValSearch(arr, 0, arr.length-1, 100);
        System.out.println("下标为"+ I); 

    } 
    public  static  int insertValSearch ( int [] ARR, int left, int right, int {findVal) 
        . The System OUT .println ( " Interpolation Find 1 " );
         // if left> rigth or not the value of findValue correct, returns -1 
        IF (left> right findVal || <ARR [ 0 ] || findVal> ARR [arr.length- . 1 ]) {
             return - . 1 ; 
        } 
        // adaptive MID 
        int MID = left + (right - left) * (findVal-arr [ left]) / (arr [right] -arr[left]);
        int midVal = arr[mid];
        if(findVal > midVal){
            return insertValSearch(arr,mid+1,right,findVal);
        }else if(findVal < midVal){
            return insertValSearch(arr,left,mid-1,findVal);
        }else {
            return mid;
        }
    }
}

D. Output: