第2章 数字之魅——快速寻找满足条件的两个数

快速寻找满足条件的两个数

问题描述

分析与解法

【解法一】

package chapter2shuzizhimei.findtwonumber;
/**
* 快速寻找满足条件的两个数
* 【解法一】
* @author DELL
*
*/
public class FindTowNumber1 {
//定义一个二元组类
public static class Tuple{
public double a;
public double b;
public Tuple(double a, double b){
this.a = a;
this.b = b;
}
}
//寻找满足条件的两个数
public static Tuple find(double a[],double sum){
int n = a.length;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(a[i]+a[j]==sum){
Tuple t = new Tuple(a[i],a[j]);
return t;
}
}
}
return null;
}
public static void main(String[] args) {
double a[] = {3,8,4,9,12,5,1};
double sum = 10;
Tuple t = find(a,sum);
System.out.println("数组中和为："+sum+"的两个数为："+t.a+" "+t.b);

}

}

数组中和为：10.0的两个数为：9.0 1.0

【解法二】

package chapter2shuzizhimei.findtwonumber;
/**
* 快速寻找满足条件的两个数
* 【解法二】
* @author DELL
*
*/
public class FindTowNumber2 {
//定义一个二元组类
public static class Tuple{
public double a;
public double b;
public Tuple(double a, double b){
this.a = a;
this.b = b;
}
}

//快速排序的一次划分
public static int partition(double a[], int first, int last) {
double temp = a[first];
int i = first, j = last;
while (i < j) {
while (i < j && a[j] >= temp) {
j--;
}
if (i < j)
a[i] = a[j];
while (i < j && a[i] <= temp) {
i++;
}
if (i < j)
a[j] = a[i];
}
a[i] = temp;
return i;
}

// 快速排序
public static void quickSort(double a[], int first, int last) {
if (first >= last)
return;
int i = partition(a, first, last);
quickSort(a, first, i - 1);
quickSort(a, i + 1, last);
}

//寻找满足条件的两个数
public static Tuple find(double a[],double sum){
int n = a.length;
quickSort(a,0,n-1);  //从小到大排序
int i,j;
for(i=0,j=n-1;i<j;){
if(a[i]+a[j]==sum){
Tuple t = new Tuple(a[i],a[j]);
return t;
}else if(a[i]+a[j]>sum){
j--;
}else{
i++;
}
}
return null;
}
public static void main(String[] args) {
double a[] = {3,8,4,9,12,5,1};
double sum = 9;
Tuple t = find(a,sum);
System.out.println("数组中和为："+sum+"的两个数为："+t.a+" "+t.b);

}

}

数组中和为：9.0的两个数为：1.0 8.0