好的,所以我需要继续获取用户输入,直到他们输入0.我需要将它们存储在一个数组中并打印出最小值。但它继续给我最后输入的数字而不是排序并获得列表中的最小值。这是我到目前为止所做的。
import java.io.*;
import java.util.*;
public class FindingMin
{
public static void main(String[] args) throws IOException
{
int[] Numbers = new int[100];
int minimum = 0;
int InputParser;
String input= "1";
try
{
InputStreamReader stream = new InputStreamReader (System.in);
BufferedReader scan = new BufferedReader(stream);
InputParser = Integer.parseInt(input);
while(InputParser != 0)
{
input = scan.readLine();
InputParser = Integer.parseInt(input);
for(int i = 0;i<Numbers.length;i++)
{
if(InputParser == 0)
{
InputParser = 0;
}
else
Numbers[i] = InputParser;
}
}
minimum = findingMin(Numbers,Numbers[0],Numbers.length-1);
System.out.println("The minimum number is "+minimum);
}
catch(NumberFormatException exception)
{
System.out.println("Please enter integers only");
}
}
public static int findingMin(int[] list, int start, int end)
{
if (start == end)
return list[start];
else
{
int Min = findingMin(list, start, end-1);
if (Min < list[end])
return list[end];
else
return Min;
}
}
}
任何建议都将不胜感激!
答案 0 :(得分:2)
那是因为你有一个额外的循环:
while(InputParser != 0)
{
input = scan.readLine();
InputParser = Integer.parseInt(input);
for(int i = 0;i<Numbers.length;i++)
{
if(InputParser == 0)
{
InputParser = 0;
}
else
Numbers[i] = InputParser; // <-- you set all numbers
// to the last input here
}
}
正确的版本是:
int i = 0;
while(InputParser != 0)
{
input = scan.readLine();
InputParser = Integer.parseInt(input);
if(InputParser == 0)
{
InputParser = 0;
}
else
{
Numbers[i] = InputParser;
i++;
}
}
此外,调用方法应为:
minimum = findingMin(Numbers,0,Numbers.length-1);
不是
minimum = findingMin(Numbers,Numbers[0],Numbers.length-1);