【发布时间】:2016-01-22 06:43:59
【问题描述】:
首先我想说我不是很有经验,如果这个问题得到了回答,我很抱歉。我一直试图找到一个答案,但一直未能。
我正在开发一个用户将数字输入数组的项目。这些数字代表不同日子的温度。天数显然是数组中的位置。我需要找到一种方法来打印温度从最低到最高而不对数组进行排序。
因此,如果用户输入 [56, 45, 67, 41, 59, 70],则表示位置 0(第 1 天)为 56 度,位置 2(第 3 天)为 67 度。我需要保持数组的位置相同,这样打印出来的时间就会保持不变。
编辑:到目前为止,我已经附上了我在项目中的代码。 HighestOrdered 方法是我不知道该做什么或从哪里开始的方法。对于我上面所说的 HighestOrdered 方法,我需要让它打印出当天的临时工(数组中的位置),我不知道该怎么做。
这是我目前的代码:
public class Weather {
public static void main(String[] args) {
// TODO Auto-generated method stub
int [] high = new int[30];
int [] low = new int[30];
Init (high);
Init(low);
LoadData(high,low);
Report(high, low);
FindAvg(high,low);
Lowest(high, low);
Highest(high,low);
}
public static void Init(int A[])
{
for(int i = 0; i < A.length; i++)
{
A[i] = 510;
}
}
public static void Report(int[] H, int[] L)
{
System.out.println("Day High Low");
for(int i = 0; i < H.length; i++)
{
System.out.println(i + " " + H[i] + " " + L[i]);
}
}
public static void LoadData(int[] H, int[] L)
{
int day = 0;
while(day < 30)
{
try {
int high = Integer.parseInt(JOptionPane.showInputDialog("please enter the high"));
H[day] = high;
} catch (NumberFormatException e) {
}
try {
int low = Integer.parseInt(JOptionPane.showInputDialog(" Please enter the low"));
L[day] = low;
} catch (NumberFormatException e) {
}
day++;
}
}
public static void FindAvg(int[] H, int[] L){
int sumHigh = 0;
int avgHigh;
int sumLow = 0;
int avgLow;
for(int i : H)
sumHigh += i;
avgHigh = sumHigh/H.length;
for(int i : L)
sumLow += i;
avgLow = sumLow/L.length;
System.out.println("The average for the high is: " + avgHigh);
System.out.println("The average for the low is: " + avgLow);
}
public static void Highest(int[] H, int[] L)
{
int highestHigh = -1000;
int dayHigh = 0;
int highestLow = -1000;
int dayLow = 0;
for(int i = 0; i < H.length; i++)
{
if(H[i] > highestHigh && H[i] != 510)
{
highestHigh = H[i];
dayHigh = i;
}
}
System.out.println("\n" + "The highest high is: " + highestHigh + " degrees." + "\n" +
"This temperature was recorded on day: " + dayHigh);
for(int i = 0; i < L.length; i++)
{
if(L[i] > highestLow && L[i] != 510)
{
highestLow = L[i];
dayLow = i;
}
}
System.out.println("\n" + "The highest low is: " + highestLow + " degrees." + "\n" +
"This temperature was recorded on day: " + dayLow);
}
public static void Lowest(int[] H, int[] L)
{
int lowestHigh = 1000;
int dayHigh = 0;
int lowestLow = 1000;
int dayLow = 0;
for(int i = 0; i < H.length; i++)
{
if(H[i] < lowestHigh)
{
lowestHigh = H[i];
dayHigh = i;
}
}
System.out.println("\n" + "The lowest high is: " + lowestHigh + " degrees." + "\n" +
"This temperature was recorded on day: " + dayHigh);
for(int i = 0; i < L.length; i++)
{
if(L[i] < lowestLow)
{
lowestLow = L[i];
dayLow = i;
}
}
System.out.println("\n" + "The lowest low is: " + lowestLow + " degrees." + "\n" +
"This temperature was recorded on day: " + dayLow);
}
public void HighestOrdered(int[] H)
{
}
}
【问题讨论】:
-
复制数组并对其进行排序。在第二个数组中,您可以跟踪索引,即进行与主数组中相同的交换。
-
不排序如何排序
-
当你说“不排序”时,你的意思是:(1)你需要保留原始的排序信息(即
[56, 45, 67, 41, 59, 70]),但是复制这个并排序就可以了;或者 (2) 这是某个班级的作业并且禁止排序?
标签: java arrays sorting methods