Question

我正在尝试编写一个简单的算法来识别ArrayList整数中第一个缺失的实数。我在编写逻辑时遇到了困难。观察我的代码snippit：

// suppose sequence is a valid sorted ArrayList

int match = 0;
int first = sequence.get(0); // sets the first value in seq. to var
int size = sequence.size(); // sets the seq size to var
for (int i = 0; i < size; i++)
{
    if (i != sequence.get(i) && i > first)
        match = i; // it is not in the sequence
}
System.out.println(match + " is not in the sequence.");

如果序列为4, 5, 8, 9，我希望match为6。相反，我得到了0。帮助

Answer 1

这是因为您的if条件要求sequence.get(i)等于i。相反，它应该是first + i，并且需要删除条件的第二部分：

if ((first + i) != sequence.get(i)) {
    match = (first + i);
    break;
}

Demo on ideone.

Answer 2

试试这个：

int max = -1;
for(int i=1; i<sequence.size(); i++) {
    if(sequence.get(i) > sequence.get(i - 1)) {
        match = i + 1;
        break;
    }
}
System.out.println(match + " is not in the sequence.");

Answer 3

支票类似于sequence.get(i+1) == sequence.get(i) + 1。如果失败，您将返回sequence.get(i) + 1。

Answer 4

import  java.util.Arrays;
import  java.util.List;
/**
   <P>{@code java FindMissingNumInSequence}</P>
 **/
public class FindMissingNumInSequence  {
   public static final void main(String[] ignored)  {
      List<Integer> ascUnqIntList = Arrays.asList(4, 5, 8, 9);

      int num = ascUnqIntList.get(0);  //Initialize to the first one
      for(int i : ascUnqIntList)  {
         if(num != i)  {
            System.out.println("First missing number in sequence: " + num);
            break;
         }
         System.out.println(i);
         num++;
      }
   }
}

输出：

[C:\java_code\]java FindMissingNumInSequence
4
5
First missing number in sequence: 6

Answer 5

这应该可以帮助您获得所需的内容，假设数组列表已经排序。

for (int i = 0; i < size - 1; i++)
{
    if (sequence.get(i+1) - sequence.get(i) > 1 ) {
        match = sequence.get(i) + 1; // it is not in the sequence
        break;
    }
}

Answer 6

Array (ans1,ans2,ans3);

Answer 7

public class MissingNumber {

    public static void main(String[] args) {
        MissingNumber num=new MissingNumber();
        int[] numb={2,4,6,10,12,14,};
        num.solution(numb);

    }
    public int solution(int[] A) {

        int j=0;
        int k=0;
        int missingNum=0;
        int[] arrayInt=A;
        int diff=0;
        for(int i=0;i<arrayInt.length-2;i++)
        {
            j=arrayInt[i+1]- arrayInt[i];

            k=arrayInt[i+2]-arrayInt[i+1];
            if(j==k){
                 diff= k;
            }
            if(j!=k)
            {
                if(j!=diff)
                { 
                    missingNum=arrayInt[i]+diff;
                }
                else if(k!=diff){
                     missingNum=arrayInt[i+1]+diff;
                }


            }

        }  
        System.out.println("Missing number"+missingNum);
        return missingNum;
    }
}

在序列中找到缺失数字的逻辑是什么？

7 个答案: