经过一夜的编程，我最终得到了一个程序，可以打印出数组的模式。或者甚至会告诉您是否没有模式（例如，如果没有输入多次发生或所有输入发生的次数相同：例如 1、1、2、2、3、3、4、4 ）。该程序的一些限制是您必须输入多个数字，并且您不能输入超过 10000 个数字或负数（如果您想输入一个负数，您只需调整所有涉及值的 for 循环[][] 数组。关于我的程序的一些很酷的事情是它会打印出每个输入发生的次数以及数组的模式。并且所有打印输出的语法都会根据信息量而变化（例如。数组的模式是 2；数组的模式是 1 和 2；数组的模式是 0、2、5 和 8)。程序中还有一个冒泡排序函数示例，供任何认为他们在他们的模式程序中需要一个排序器功能。希望这会有所帮助，我包含了很多伪代码来帮助那些看不到我的逻辑在整个程序中如何进展的人。（仅供参考：它是 java，并在 BlueJ 中编译）
import java.util.Scanner;
public class Mode
{
public static void main (String args [])
{
Scanner scan = new Scanner(System.in);
int MAX_INPUTS = 10000; boolean flag = false;
System.out.print ("Input the size of your array: ");
int size; // How many nubers will be in the user array
do
{
size = scan.nextInt();
if (size == 1)
{
System.out.print ("\nError. You must enter a number more than 1.\n\n");
continue;
}
else if (size > MAX_INPUTS || size < 0)
{
System.out.print ("\nError. You muste enter a number less than " + MAX_INPUTS + " or greater than 0\n\n");
continue;
}
else
flag = true; // a ligit answer has been entered.
}
while (flag != true);
int array[] = new int[size], values[][] = new int[2][MAX_INPUTS + 1], ticks = 0;
System.out.print ("\nNow input the numbers for your array.\n\n");
/* Taking inputs from the user */
while (ticks < size)
{
System.out.print ("Number " + (ticks + 1) + ": ");
array[ticks] = scan.nextInt();
if (array[ticks] > MAX_INPUTS || array[ticks] < 0)
{
System.out.print ("\nError. Number cannot be greater than " + MAX_INPUTS + " or less than 0\n\n");
continue;
}
++ticks;
}
/*
* values[][] array will hold the info for how many times numbers 0 - 10000 appear in array[]. Column 0 will hold numbers from 0 -1000, and column 1 will hold the number of
* of repititions the number in column 0 occured in the users inputed array.
*/
for (int i = 0; i < MAX_INPUTS; ++i) // Initalize Column zero with numbers starting at zeor, and ending and MAX_INPUTS.
values[0][i] = i;
for (int i = 0; i < size; ++i) // Find the repatitions of the numbers in array[] that correspond to the number in column zere of values[][].
for (int j = 0; j < MAX_INPUTS; ++j)
if (array[i] == j)
++values[1][j];
sort (array, size);
System.out.print ("\n\nHere are the numbers you entered.\n\n"); // show the values the user entered in ascending order.
for (int i = 0; i < size; ++i)
{
if (i == size - 1) // the last inputed number
System.out.print (array[i]); // don't allow an extra comma.
else
System.out.print (array[i] + ", ");
}
// Show the user how many times each of the values he/she entered occured.
System.out.print ("\n\nThis is the amount of times each of the values you entered occured:\n");
for (int i = 0; i < MAX_INPUTS; ++i)
{
if (values[1][i] == 1)
System.out.print (i + " was entered " + values[1][i] + " time\n"); // avoid: 2 was entered 1 times
else if (values[1][i] != 0)
System.out.print (i + " was entered " + values[1][i] + " times\n"); // avoid: 2 was entered 2 time
}
/* -------------------------------------------------------------------- | Finding the Mode/Modes | -------------------------------------------------------------------- */
/* The process begins with creating a second array that is the exactly the same as the values[][] (First for loop). Then I sort the duplicate[] array to find the mode
* (highest number in the duplicate[]/values[][] arrays. Int max is then assigned the highest number. Remembering that the values[][] array: column 0 contains numbers ranging
* from 1 to 10000, it keeps track of where the numbers in column were originally located, in which you can compare to the duplicate array which is sorted. Then I can set
* up a flag that tells you whether there is more than one mode. If so, the printing of these modes will look neater and the grammar can be changed accordingly.
*/
int duplicate[] = new int [10001], mode[] = new int [size], max, mode_counter = 0;
boolean multi_mode = false, all_same;
for (int i = 0; i < MAX_INPUTS; ++i)
duplicate[i] = values[1][i]; // copy values array.
sort (duplicate, MAX_INPUTS);
max = duplicate[MAX_INPUTS - 1]; // the last number in the sorted array is the greatest.
all_same = test (duplicate, MAX_INPUTS, size, max); // this is the test to see if all the numbers in the user array occured the same amount of times.
int c = 0; // a counter
/* The mode of the user inputed array will be recorded in the values array. The sort of the duplicate array told me what was the higest number in that array. Now I can
* see where that highest number used to be in the original values array and recored the corresponding number in the column zero, which was only filled with numbers 0 -
* 10000. Thus telling me the mode/modes.
*/
for (int i = 0; i < MAX_INPUTS; ++i)
{
if (values[1][i] == max)
{
mode[c++] = values[0][i];
++mode_counter;
}
}
if (mode[1] != 0) //mode[0] (the first cell, has a number stored from the last for loop. If the second cell has a number other than zero, that tells me there is more than 1 mode.
multi_mode = true;
if (multi_mode == false)
System.out.print ("\nThe mode of your array is " + mode[0]); // For correct grammer.
else if (all_same == true)
System.out.print ("\nAll of the numbers entered appeared the same amount of times. "); // See the boolean function for more details
else // If here there is more than one mode.
{
System.out.print ("\nThe modes of yoru array are ");
for (int i = 0; i < mode_counter; ++i)
{
if (mode_counter > 2 && i == (mode_counter - 1)) // If there is more than two modes and the final mode is to be printed.
System.out.print ("& " + mode[i]);
else if (mode_counter == 2)
{ // This is true if there is two modes. The else clause will print the first number, and this will print the amper sign and the second mode.
System.out.print (mode[0] + " & " + mode[1]);
break;
}
else
System.out.print (mode[i] + ", ");
}
}
}
public static void sort (int list[], int max) // Its the bubble sort if you're wondering.
{
int place, count, temp;
for (place = 0; place < max; ++place)
for (count = max - 1; count > place; --count)
if (list[count - 1] > list[count])
{
temp = list[count-1];
list[count - 1] = list[count];
list[count] = temp;
}
}
/* The test to see if there isn't a mode. If the amount of the mode number is the same as the amount of numbers there are in the array is true, or if the size entered by the
* user (input) modulo the mode value is equal to zero (say, all the numbers in an array of ten were entered twice: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5). */
public static boolean test (int list[], int limit, int input, int max)
{
int counter = 0, anti_counter = 0;
for (int i = 0; i < limit; ++i)
if (list[i] == max)
++counter; // count the potential modes
else if (list[i] !=0 && list[i] != max)
++anti_counter; // count every thing else except zeros.
if (counter == input || (input % max == 0 && anti_counter == 0) )
return true;
else
return false;
}
}