二分查找法的问题

[quietwater]

你随机生成一个double类型数据的数组，长度为100，对它进行排序，然

后交互式地提出用户输入数字，最后显示数组中与它最接近的一个大

数和一个小数。使用二分查找法在排序后的数组中进行查找定位。

请给出源代码和必要的说明，编程思路。

[hayai]

public class Tester
{
private final static int capacity = 100;
private static double[] ary = new double[capacity];

public static void main(String[] args)
{
generateArray();
sortArray();
double num = randomNumber();
int position = searchArray(num, 0, capacity);
showResult(position, num);
}

private final static int max = 1000;
private static double randomNumber()
{
return Math.random() * 1000;
}

private static void generateArray()
{
for (int i = 0; i < capacity; i++)
ary[i] = randomNumber();
}

private static void sortArray()
{ // Simple Sort Algorithm
for (int i = 0; i < capacity - 1; i++)
for (int n = i + 1; n < capacity; n++)
if (ary[i] > ary[n])
{
double temp = ary[i];
ary[i] = ary[n];
ary[n] = temp;
}
}

private static int searchArray(double num, int left, int right)
{ // Binary Search Algorithm
if (right <= left)
return left;

int point = (right - left) / 2 + left;
if (ary[point] == num)
return point;

if (ary[point] < num)
return searchArray(num, point + 1, right);
else
return searchArray(num, left, point - 1);
}

private static void showResult(int pos, double num)
{
System.out.println("Searched number: " + num);
double num1 = ary[pos];
if (num1 == num)
{
System.out.println(num1);
return;
}

if (num1 < num)
{
System.out.println(ary[pos]);
if (++pos < capacity)
System.out.println(ary[pos]);
}
if (num1 > num)
{
if (--pos >= 0)
System.out.println(ary[pos]);
System.out.println(ary[pos+1]);
}
}
}

[方工]

将下面的有关int的改为double就可以了：
/** BiSearch.java
* create an array of 100 integers, sort it, binary search for a given value;
*/
import java.io.*;
import java.util.Random;
public class BiSearch
{
public static void swap(int[] data, int i, int j){
int t = data[i];
data[i] = data[j];
data[j] = t;
}
public static int find(int[] data, int x){
int low = 0, high = data.length-1;
int mid;
while(low<=high){
mid = (low + high)/2;
if(data[mid]==x) return mid;
else if(data[mid]>x) high = mid-1;
else low = mid+1;
}
return -1;
}

public static void dump(int[] data){
System.out.println("数组内容:");
for(int i=0; i<data.length; i++){
System.out.print(data[i] + ", ");
}
System.out.println("");
}
public static void selectionSort(int[] data, int n){
int unSorted = n;
int index;
int max;
while(unSorted > 0){
max = 0;
for(index = 0; index< unSorted; index++){
if(data[max] < data[index]) max = index;
}
swap(data, max, unSorted-1);
unSorted --;
}
}

public static void main(String[] args)
{
final int SIZE = 100;
int [] arr= new int[SIZE];
Random r = new Random(0);

for(int i=0; i<SIZE; i++){
arr[i] = r.nextInt()%1000;
}
dump(arr);
selectionSort(arr, SIZE);
dump(arr);
String s;
try{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.println("请输入查找的数(按回车结束):");
while((s=in.readLine()).length()!=0){
int x = Integer.parseInt(s);
int pos;
pos = find(arr, x);
if(pos>=0){
System.out.println("位置:" + pos + ", 数值:" + arr[pos]);
if(pos>0)
System.out.println("前一个值:" + arr[pos-1]);
if(pos<arr.length-1)
System.out.println("下一个值:" + arr[pos+1]);
}
else
System.out.println("没有找到:" + x);
System.out.println("请输入查找的数(按回车结束):");
}
}catch(Exception e){
System.out.println(e);
System.exit(0);
}
}
}

[quietwater]

hayai(生命树)和onestab(一刺)的方法比较我比较喜欢onestab的方法
因为效率比较好，hayai的方法使用递归的方法效率要差很多。
感谢你们的帮助，以下是我的程序，希望大家取长补短，共同进步。



import java.io.*;
import java.util.*;
class nuts1_6{
public static void main(String[] args)throws IOException{
double[] d=new double[100];
d=CreateDoubleArray.create(100);
Arrays.sort(d);
for(int i=0;i<d.length;i++){
System.out.print(d[i]+" ");
if(i%4==0)
System.out.println();
}

BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
for(;;){
int i=0;
System.out.print(">");
String line=in.readLine();
if((line==null)||(line.equals("quit"))) break;
try{
double j=Double.parseDouble(line);
i=CreateDoubleArray.FindDouble(d,j);
}
catch(NumberFormatException e){
System.out.println("please input a double");
continue;
}
if(i==-1)
System.out.println("Can't find");
else
System.out.println(i);
}


}
}

[quietwater]

以下是一个更通用的二分法查找的源程序：import java.util.*;
class CreateDoubleArray{
static double[] create(int n){
double[] d=new double[n];
for(int i=0;i<n;i++)
d[i]=Math.random()*100;
return d;
}
static int FindDouble(double[] d,double j){
int left=0;
int right=d.length-1;
if(j==d[left])
return left;
if(j==d[right])
return right;
for(;;){
if(j==d[(left+right)/2]){
return (left+right)/2;
}
else if(j<d[(left+right)/2])
right=(left+right)/2;
else if(j>d[(left+right)/2])
left=(left+right)/2;
if(left==(left+right)/2)
return -1;
}
}
}


import java.io.*;
import java.util.*;
class nuts1_6{
public static void main(String[] args)throws IOException{
double[] d=new double[100];
d=CreateDoubleArray.create(100);
Arrays.sort(d);
for(int i=0;i<d.length;i++){
System.out.print(d[i]+" ");
if(i%4==0)
System.out.println();
}

BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
for(;;){
int i=0;
System.out.print(">");
String line=in.readLine();
if((line==null)||(line.equals("quit"))) break;
try{
double j=Double.parseDouble(line);
i=CreateDoubleArray.FindDouble(d,j);
}
catch(NumberFormatException e){
System.out.println("please input a double");
continue;
}
if(i==-1)
System.out.println("Can't find");
else
System.out.println(i);
}


}
}

[quietwater]

hayai(生命树)和onestab(一刺)的方法比较我比较喜欢onestab的方法
因为效率比较好，hayai的方法使用递归的方法效率要差很多。
感谢你们的帮助，以下是我的程序，希望大家取长补短，共同进步。


import java.io.*;
class nuts1_5{

public static void main(String[] args)throws IOException{
double[] nums=new double[100];
for(int i=0;i<nums.length;i++)
nums[i]=Math.random()*100;
Arrays.sort(nums);
for(int i=0;i<nums.length;i++){
System.out.print(nums[i]+" ");
if(i%4==0)
System.out.println();
}
BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
for(;;){
System.out.print(">");
String line=in.readLine();
if((line==null)||(line.equals("quit"))) break;
double j=Double.parseDouble(line);
int left=0,right;
right=nums.length-1;
if(j==nums[0]){
System.out.println(nums[0]);
continue;
}
if(j==nums[right]){
System.out.println(nums[right]);
continue;
}
for(;;){
if(j==nums[(left+right)/2]){
System.out.println(nums[(left+right)/2]);
break;
}
else if(j<nums[(left+right)/2])
right=(left+right)/2;
else if(j>nums[(left+right)/2])
left=(left+right)/2;
if(left==(left+right)/2){
System.out.println(nums[left]+" "+nums[right]);
break;
}

}
}
}
}