我想获取我生成的随机值数组，并在相同数字的最长运行之外用括号打印上述数组。 例如，如果数组是[0,1,1,1,2,4,7,4]，我想收到0（111）2474作为输出。
到目前为止，这是我的代码。
import java.util.Random;
import java.util.Arrays;
/**
* Write a description of class ArrayRunner1 here.
*
* @author Ibrahim Khan
* @version (a version number or a date)
*/
public class ArrayRunner1 {
/**
* This method will generate my random numbers for my array.
* @param min minimum random value wanted
* @param max maximum random value wanted
* @return randomNum a random number between 1 and 6 inclusive
*/
public static int randInt(int min, int max) {
Random rand = new Random();
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
public static void main(String[] args) {
System.out.println("\f");
//Part 1 - Generate a random array of length 40 with random 1-6 inclusive
int[] array1 = new int[40];
for (int i = 0; i < array1.length; i++) {
array1[i] = randInt(1, 6);
}
System.out.println(Arrays.toString(array1));
//Counts and RETURN: reports how many times each number is present
int counter1 = 0;
int counter2 = 0;
int counter3 = 0;
int counter4 = 0;
int counter5 = 0;
int counter6 = 0;
for (int i = 0; i < array1.length; i++) {
if (array1[i] == 1) {
counter1++;
}
if (array1[i] == 2) {
counter2++;
}
if (array1[i] == 3) {
counter3++;
}
if (array1[i] == 4) {
counter4++;
}
if (array1[i] == 5) {
counter5++;
}
if (array1[i] == 6) {
counter6++;
}
}
System.out.println("There are " + counter1 + " ones.");
System.out.println("There are " + counter2 + " twos.");
System.out.println("There are " + counter3 + " threes.");
System.out.println("There are " + counter4 + " fours.");
System.out.println("There are " + counter5 + " fives.");
System.out.println("There are " + counter6 + " sixes.");
//Counts the longest run of the same number. A run continues only when consecutive numbers have the same value.
//RETURN: The repeated number and the length of the run is then printed
int counter = 1;
int runMax = 1;
int runMin = 0;
int variableNum = 0;
int startCounter = 0;
int endCounter = 0;
for (int i = 0; i < array1.length - 1; i++) {
if (array1[i] == array1[i + 1]) {
counter++;
if (counter >= runMax {
runMax = counter;
runMin = i - counter + 1;
variableNum = array1[i];
startCounter = i - counter + 2;
endCounter = i + counter - 1;
}
} else {
counter = 1;
}
}
System.out.println("The longest run is " + runMax + " times and the number is " + variableNum + ". ");
System.out.println("The run starts at " + startCounter + " and ends at " + endCounter);
//Prints the array with parentheses outside the longest run, if there is more than one max run, use the last one.
}
}
答案 0 :(得分：0)
试试这段代码：
import java.util.Arrays;
import java.util.Random;
public class Snippet {
/**
* This method will generate my random numbers for my array.
*
* @param min
* minimum random value wanted
* @param max
* maximum random value wanted
* @return randomNum a random number between 1 and 6 inclusive
*/
public static int randInt(int min, int max) {
Random rand = new Random();
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
public static void main(String[] args) {
System.out.println("\f");
// Part 1 - Generate a random array of length 40 with random 1-6
// inclusive
int[] array1 = new int[40];
for (int i = 0; i < array1.length; i++) {
array1[i] = randInt(1, 6);
}
System.out.println(Arrays.toString(array1));
// Counts and RETURN: reports how many times each number is present
int counter1 = 0;
int counter2 = 0;
int counter3 = 0;
int counter4 = 0;
int counter5 = 0;
int counter6 = 0;
for (int i = 0; i < array1.length; i++) {
if (array1[i] == 1) {
counter1++;
}
if (array1[i] == 2) {
counter2++;
}
if (array1[i] == 3) {
counter3++;
}
if (array1[i] == 4) {
counter4++;
}
if (array1[i] == 5) {
counter5++;
}
if (array1[i] == 6) {
counter6++;
}
}
System.out.println("There are " + counter1 + " ones.");
System.out.println("There are " + counter2 + " twos.");
System.out.println("There are " + counter3 + " threes.");
System.out.println("There are " + counter4 + " fours.");
System.out.println("There are " + counter5 + " fives.");
System.out.println("There are " + counter6 + " sixes.");
// Counts the longest run of the same number. A run continues only when
// consecutive numbers have the same value.
// RETURN: The repeated number and the length of the run is then printed
int counter = 1;
int runMax = 0;
int runMin = 0;
int variableNum = 0;
int startCounter = 0;
int endCounter = 0;
for (int i = 0; i < array1.length - 1; i++) {
if (array1[i] == array1[i + 1]) {
counter++;
if (counter >= runMax) {
runMax = counter;
startCounter = i - counter +2;
// runMin = i-counter+1;
variableNum = array1[i];
endCounter = i+1;
}
} else {
counter = 1;
}
}
System.out.println("The longest run is " + runMax
+ " times and the number is " + variableNum + ". ");
System.out.println("The run starts at " + startCounter
+ " and ends at " + endCounter);
for (int i = 0; i < array1.length; i++) {
if (i==startCounter) {
System.out.print("(");
}
System.out.print(array1[i]);
if (i==endCounter) {
System.out.print(")");
}
}
System.out.println();
// Prints the array with parentheses outside the longest run, if there
// is more than one max run, use the last one.
}
}
答案 1 :(得分：0)
好。我想我有这个。第一个答案很接近，但如果您运行该程序几次，就会发现问题。在上面的代码中有一个逻辑错误，但我有一个解决方法。我想你是如何得到endCounter的。这似乎很奇怪。但据我所知，我得到了该计划。试试吧。我已经多次运行它似乎一致。
import java.util.Random; import java.util.Arrays;
/ ** *在这里写一个ArrayRunner1类的描述。 * * @author Ibrahim Khan * @version（版本号或日期） * / 公共类ArrayRunner1 {
/**
* This method will generate my random numbers for my array.
* @param min minimum random value wanted
* @param max maximum random value wanted
* @return randomNum a random number between 1 and 6 inclusive
*/
public static int randInt(int min, int max) {
Random rand = new Random();
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
public static void main(String[] args) {
System.out.println("\f");
//Part 1 - Generate a random array of length 40 with random 1-6 inclusive
int[] array1 = new int[40];
for (int i = 0; i < array1.length; i++) {
array1[i] = randInt(1, 6);
}
System.out.println(Arrays.toString(array1));
//Counts and RETURN: reports how many times each number is present
int counter1 = 0;
int counter2 = 0;
int counter3 = 0;
int counter4 = 0;
int counter5 = 0;
int counter6 = 0;
for (int i = 0; i < array1.length; i++) {
if (array1[i] == 1) {
counter1++;
}
if (array1[i] == 2) {
counter2++;
}
if (array1[i] == 3) {
counter3++;
}
if (array1[i] == 4) {
counter4++;
}
if (array1[i] == 5) {
counter5++;
}
if (array1[i] == 6) {
counter6++;
}
}
System.out.println("There are " + counter1 + " ones.");
System.out.println("There are " + counter2 + " twos.");
System.out.println("There are " + counter3 + " threes.");
System.out.println("There are " + counter4 + " fours.");
System.out.println("There are " + counter5 + " fives.");
System.out.println("There are " + counter6 + " sixes.");
//Counts the longest run of the same number. A run continues only when consecutive numbers have the same value.
//RETURN: The repeated number and the length of the run is then printed
int counter = 1;
int runMax = 1;
int runMin = 0;
int variableNum = 0;
int startCounter = 0;
int endCounter = 0;
for (int i = 0; i < array1.length - 1; i++) {
if (array1[i] == array1[i + 1]) {
counter++;
if (counter >= runMax ){
runMax = counter;
runMin = i - counter ;// was plus one I cahnged this.
variableNum = array1[i];
startCounter = i - counter + 2;
endCounter = i + counter -1;
}
} else {
counter = 1;
}
}
System.out.println("The longest run is " + runMax + " times and the number is " + variableNum + ". ");
System.out.println("The run starts at " + startCounter + " and ends at " + endCounter);
//Prints the array with parentheses outside the longest run, if there is more than one max run, use the last one.
String output = "";// added this
for(int x = 0; x < array1.length; x++)
{
if( x == startCounter)
{
output += "("+array1[x];
}
else if( x == startCounter + runMax )
{
else if( x == startCounter + runMax )
{
if(x == array1.length-1)
{
output += ")";
}
else
{
output += ")"+array1[x];
}
}
else
{
output += array1[x];
}
}
System.out.print("\n"+output);
}
}
答案 2 :(得分：0)
这是一个更短，更通用的解决方案。此方法采用任意数组，并在最长的数字周围打印括号。如果有两个相同长度的跑步，它会围绕第一个打印。
public String makeString(int[] ints) {
if (ints.length == 0) return ""; // Quit early if there's nothing to do.
// Initialize variables.
int lastNumber = ints[0];
// We keep track of the all time best run. Defaults to first int found.
int bestStart = 0;
int bestRun = 1;
// ... as well as the current run.
int currentStart = 0;
int currentRun = 1;
String s = ""+ints[0];
// Starting from the second int, we check if the current run is continuing.
for (int i = 1; i < ints.length; i++) {
int current = ints[i];
// If the current run continues, we update currentStart/currentRun, else we reset it.
if (current == lastNumber) {
currentRun++;
} else {
currentStart = i;
currentRun = 1;
}
// Now we check if the currentRun is better than the best.
// If so, we update bestStart/bestRun.
if (currentRun > bestRun) {
bestStart = currentStart;
bestRun = currentRun;
}
lastNumber = current;
s += current;
}
// Now that we've found it, we insert parenthesis aaaaaaand we're done!
return s.substring(0, bestStart)
+"("+s.substring(bestStart, bestStart+bestRun)+")"
+s.substring(bestStart+bestRun);
}