Question

给定一组int，如果值3在数组中出现正好3次，则返回true，并且没有3个是彼此相邻的。

haveThree({3, 1, 3, 1, 3}) → true
haveThree({3, 1, 3, 3}) → false
haveThree({3, 4, 3, 3, 4}) → false

public boolean haveThree(int[] nums)
{
      int count = 0;
          boolean isPerv3 = false;
          for(int i = 0 ; i < nums.length && count <= 3; i++)
          {
            if(nums[i] == 3)
            {
                if(isPerv3)
                    return false;
                else
                {
                    count++;
                    isPerv3 = true;
                }
            }
            else
                isPerv3 = false;
          }
          return (count == 3);
}

Answer 1

使用ListIterator对象对于这种情况是理想的。这样，您可以将List中的下一个值与之前的值进行比较。您可以保留一个计数器来检查并查看是否有三个连续3个。

import java.util.ArrayList;
import java.util.ListIterator;

public class Test {

    public static void main(String[] args) {

        int[] test1 = { 3, 2, 5, 11, 11, 11, 233, 22, 3, 3, 3, 5, 5, 5 };   // false
        int[] test2 = { 3, 3, 3, 8, 3, 11, 233, 22, 3, 3, 3, 5, 5, 5 };     // false
        int[] test3 = { 3, 2, 5, 8, 3, 11, 233, 22, 3, 3, 4, 3, 3, 3 };     // false
        int[] test4 = { 3, 2, 5, 8, 3, 11, 233, 22, 3, 3, 5, 5, 5 };        // true
        System.out.println(haveThree(test1));
        System.out.println(haveThree(test2));
        System.out.println(haveThree(test3));
        System.out.println(haveThree(test4));

    }

    public static boolean haveThree(int[] nums) {

        int totalNumberOfThrees = 0;
        int threeInSuccessionCount = 0;
        boolean result = false;

        // convert to ArrayList
        ArrayList<Integer> arrayList = new ArrayList<Integer>();
        for (int i = 0; i < nums.length; i++) {
            arrayList.add(nums[i]);
        }

        // use ListIterator to compare next value with previous value
        ListIterator<Integer> iterator = arrayList.listIterator();

        while (iterator.hasNext()) {
            Integer next = iterator.next();
            Integer previous = iterator.previous();
            // counts total number of 3's
            if (next.intValue() == 3) {
                totalNumberOfThrees++;
            }
            // if next and previous values == 3 then increase count by 1
            if (next.intValue() == 3 && previous.intValue() == 3) {
                threeInSuccessionCount++;
            } 
            // if next and previous values != 3 then reset counter
            else {
                threeInSuccessionCount = 0;
            }
            // if there are three consecutive 3's set threeInSuccessionCount == 3 
            // and then break from loop 
            if (threeInSuccessionCount == 3) {
                break;
            }
            iterator.next();
        }

        if (threeInSuccessionCount != 3 && totalNumberOfThrees >= 3) {
            result = true;

        } else {
            result = false;
        }

        return result;
    }
}

输出：

false
false
false
true

修改

要回答您的问题＆＃39; 如果不声明为静态，我如何调用（在主要方法内）haveThree（）？＆＃39;，您可以在main方法中声明一个新的Test实例，然后使用它来调用haveThree（）。

像这样：

public static void main(String[] args) { int[] test1 = { 3, 2, 5, 11, 11, 11, 233, 22, 3, 3, 3, 5, 5, 5 }; int[] test2 = { 3, 3, 3, 8, 3, 11, 233, 22, 3, 3, 3, 5, 5, 5 }; int[] test3 = { 3, 2, 5, 8, 3, 11, 233, 22, 3, 3, 4, 3, 3, 3 }; int[] test4 = { 3, 2, 5, 8, 3, 11, 233, 22, 3, 3, 5, 5, 5 }; Test test = new Test(); System.out.println(test.haveThree(test1)); System.out.println(test.haveThree(test2)); System.out.println(test.haveThree(test3)); System.out.println(test.haveThree(test4)); }

我想在ecllipse中运行此代码，它在codingBat.com上运行良好

1 个答案: