Divide-and-Conquer(P) if |P|≤n0 then return(ADHOC(P)) //将P分解为较小的子问题 P1 ,P2 ,...,Pk for i←1 to k do yi ← Divide-and-Conquer(Pi) △ 递归解决Pi T ← MERGE(y1,y2,...,yk) △ 合并子问题 return(T)
import java.util.Scanner;
public class BinarySearch {
public static int BinarySearch (int[] a,int x,int n) {
int left = 0;
int right = n - 1;
while(left <= right) {
int middle = (left + right) / 2;
if(x == a[middle]) return middle;
if(x >= a[middle]) left = middle + 1;
else right = middle - 1;
}
return -1;
}
public static void main(String args[]) {
System.out.println("编程素材网测试结果:");
int[] a = new int[10];
for(int i = 0; i < a.length; i++) {
a[i] = i+1;
System.out.print(a[i] + " ");
}
System.out.println();
System.out.println("请输入你要查询的数:");
Scanner sc = new Scanner(System.in);
int b = sc.nextInt();
int num = BinarySearch(a, b, a.length) + 1;
System.out.println("要查找的数在第" + num + "个位置");
}
}
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