1. Description

Given a list of 

n numbers in non-decreasing order A={a1,a2,⋯,an} such that a1≤a2≤⋯≤an and a number x, the objective is to determine if x is present in the list A

2. Algorithm

Algorithm 1.

• Linear Search Algorithm
  • Iterate through n numbers to find the x.

Algorithm 2.

• Binary Search Algorithm.

3. Time complexity

Algorithm 1.

O(n)

Algorithm 2.

O(logn)

4. Implementation

Algorithm 1.

public class LinearSearch {public static void Result(int Seq[],int Len,int x){for(int i=0;i<Len;i++){if(x==Seq[i]){System.out.println(x+" is present at seq["+i+"]");break;}else if(x!=Seq[i]&&i==Len-1){System.out.println(x+" is not present in seq");}}}public static void main(String[] args){int Seq[]={4,33,40,44,45,45,55,60,99};int Len=Seq.length;Result(Seq,Len,55);	Result(Seq,Len,70);}
}

Result of Algorithm1:

Algorithm 2-Implementation1

public class BinarySearch1 {public static void BinarySearch(int Seq[],int x,int left,int right){		int mid=(left+right)/2;if(left==right-1&&x!=Seq[left]&&x!=Seq[right]) //如果只剩两个数，两个数都不是x，那么x不在数组Seq中{System.out.println(x+" is not present in Seq");}else if(x==Seq[mid]){System.out.println(x+" is present at Seq["+mid+"]");}else if(x>Seq[mid]){	System.out.println(x+" is between seq["+mid+"] and Seq["+right+"]");left=mid;BinarySearch(Seq,x,left,right);}else if(x<Seq[mid]){System.out.println(x+" is between seq["+left+"] and Seq["+mid+"]");right=mid;BinarySearch(Seq,x,left,right);}}public static void main(String args[]){int Seq[]={1,2,3,3,8,9,11,56,89};int left=0, right=Seq.length-1;System.out.println("Test whether 56 is present in seq:");BinarySearch(Seq,56,left,right);System.out.println("Test whether 7 is present in seq:");BinarySearch(Seq,7,left,right);}
}

Algorithm 2-Implementation2

public class BinarySearch2 {public static void BinarySearch(int Seq[],int x){int left=0;int right=Seq.length-1;int mid=(left+right)/2;while(true){if(x==Seq[mid]){System.out.println(x+" is present at Seq["+mid+"]");break;}else if(x>Seq[mid]){System.out.println(x+" is between seq["+mid+"] and Seq["  +right+"]");  left=mid+1;if(left>=right){System.out.println(x+" is not in seq");break;}mid=(left+right)/2;}else if(x<Seq[mid]){System.out.println(x+" is between seq["+left+"] and Seq["  +mid+"]");  right=mid-1; if(left>=right){System.out.println(x+" is not in seq");break;}mid=(left+right)/2;}}}public static void main(String[] args){int Seq[]={4,33,40,44,45,45,55,60,99};   System.out.println("Test whether 56 is present in seq:");  BinarySearch(Seq,55);  System.out.println("Test whether 7 is present in seq:");  BinarySearch(Seq,70);  }
}

Result of Both implementation1 and implementation2: