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如何从int []数组(Java)打印最长的数字序列

[英]How to print the longest sequence of numbers from int[] array (Java)

I'm a beginner in programming and I need to print the longest sequence of numbers from int[] array. 我是编程的初学者,我需要从int []数组中打印最长的数字序列。 For example, if we have: 例如,如果我们有:

int[] numbers = {1, 3, 3, 5, 5, 5, 5, 5, 5, 6, 0, 12, 2, 2, 2, 12, 0};

Result should be: 结果应该是:

String result = "5, 5, 5, 5, 5, 5";

I wrote some bad code that don't work, but maybe it will give you some ideas. 我写了一些不起作用的坏代码,但也许会给你一些想法。

public String findLargestSequence(int[] numbers) {
        int bestStart = 0;
        int curStart = 0;
        int bestLength = 1;
        int curLength = 1;
        for (int i = 1; i < numbers.length; i++) {
            if (numbers[i] > numbers[i - 1]) {
                curLength++;
                if (curLength > bestLength) {
                    bestStart = curStart;
                    bestLength = curLength;
                }
            } else {
                curStart = i;
                curLength = 1;
            }
        }
        List<String> identical = new ArrayList<>();
        for (int i = 0; i < bestLength; i++) {
            identical.add(String.valueOf(numbers[bestStart + i]));
        }
        return Joiner.on(", ").join(identical);
    }

Update. 更新。 Thanks to @phatfingers, I've found problem: (numbers[i] > numbers[i - 1]) should be (numbers[i] == numbers[i - 1]) . 感谢@phatfingers,我发现问题: (numbers[i] > numbers[i - 1])应该是(numbers[i] == numbers[i - 1]) But still there is one another problem. 但还是有另一个问题。 If we have something like: 如果我们有类似的东西:

int[] numbers = {1, 2, 3, 3, 4, 4};

The result of it is: 结果是:

"3, 3"

I think in this case, we could: 我想在这种情况下,我们可以:

1) say, that we don't have any one longest sequence OR 1)说,我们没有任何一个最长的序列OR

2) show all that sequences, like: 2)显示所有序列,如:

String result = "Founded sequences: " + sequence1 + ", " + sequence2;

3) do nothing with that code above. 3)对上面的代码不做任何处理。

What would you do? 你会怎么做?

this show the max occurrence, you can also count it and print them 这显示最大值,您也可以计算并打印它们

public static int consecutive(int[] array) {
        if (array.length <= 1) {
            return array.length;
        }    
        int maxRun = 0;
        for (int i = 1; i < array.length; i++) {
            int thisRun = 1;
            while (i < array.length && array[i - 1] + 1 == array[i]) {
                thisRun++;
                i++;
            }
            if (maxRun < thisRun) { // checking geater occurance
                maxRun = thisRun;
            }
        }
        return maxRun;
    }

You have to handle 4 cases in the algorithm that you can divide in two parts: 您必须处理算法中的4个案例,您可以分为两部分:

Setting the state of the current serie : 设置当前系列的状态:

  • increment the current serie if it grows 如果它增长,增加当前系列
  • reinit the current serie when it changes 当它变化时重新启动当前系列

Setting the state of the max serie : 设置最大系列的状态:

  • increment the max serie if it grows 如果它增长,增加最大系列
  • reinit the max serie when it changes 当它发生变化时重新启动最大系列

In the actual code these conditions are not respected in the loop. 在实际代码中,循环中不遵守这些条件。

I commented two logic errors to illustrate the problem : 我评论了两个逻辑错误来说明问题:

  if (numbers[i] > numbers[i - 1]) {
    // error : you don't increment the current serie when it grows
    // because the condition if true only if the the current value is 
    // superior to the previous one
    curLength++;
    if (curLength > bestLength) {
        bestStart = curStart;
        bestLength = curLength;
    }
  } 
   // error : you don't reinit the current serie only when it changes
   // because numbers[i] <= numbers[i - 1] is not true if the new number is   
   // superior to the previous one while it is a change
   else {     
    curStart = i;
    curLength = 1;
   }
 }

Here is the proposed code that handles the 4 conditions two by two: 以下是两个两个处理4个条件的建议代码:

public static String findLargestSequence(int[] numbers) {

    // variables to maintain the max serie found and the number associated
    int sizeMaxSerieFound = 0;
    int numberMaxFound = numbers[0];

    // variables to maintain the current serie found and the number
    // associated
    int sizeMaxCurrentSerie = 0;
    int numberCurrentSerie = numbers[0];

    boolean isMaxSerieIsTheCurrent = true;

    for (int i = 0; i < numbers.length; i++) {
        int currentNumber = numbers[i];

        // FIRST STEP : set the state of the current serie

        // increment the current serie if it grows or for the first
        // iteration
        if (currentNumber == numberCurrentSerie) {
            sizeMaxCurrentSerie++;
        }
        // else we reinit to 1 the current serie
        else {
            sizeMaxCurrentSerie = 1;
            numberCurrentSerie = currentNumber;
            isMaxSerieIsTheCurrent = false;
        }

        // SECOND STEP : set the state of the max serie

        // we increment the current number of the actual max serie
        if (currentNumber == numberMaxFound && isMaxSerieIsTheCurrent) {
            sizeMaxSerieFound++;
        }

        // we reinit the serie because we have found a new greater serie
        else if (currentNumber != numberMaxFound && sizeMaxCurrentSerie > sizeMaxSerieFound) {
            sizeMaxSerieFound = sizeMaxCurrentSerie;
            numberMaxFound = currentNumber;
            isMaxSerieIsTheCurrent = true;
        }

    }

    List<String> identical = new ArrayList<>();
    for (int i = 0; i < sizeMaxSerieFound; i++) {
        identical.add(String.valueOf(numberMaxFound));
    }
    return Joiner.on(", ").join(identical);
}

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