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]) //如果只剩兩個數，兩個數都不是x，那么x不在數組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:

總結

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