C程序在一系列数字中找到缺失的整数

Question

You are given a sequence of n-1 distinct positive integers, all of which are less than or equal to a integer 'n'. You have to find the integer that is missing from the range [1,2,...,n]. Solve the question without using arrays.

Input Format: One line containing the integer 'n' where 2<=n<=10,000 First line is followed by a sequence of 'n-1' distinct positive integers. Note that the sequence may not be in any particular order.

I got code by using arrays

#include<stdio.h>
int main()
{
 int i,j,n[9999],m,t;
 scanf("%d",&m);
 for(i=1;i<m;i++)
  {
   scanf("%d",&n[i]);
  }
 for(i=1;i<m;i++)
  {
   for(j=1;j<i;j++)
    {
      if(n[j]>n[j+1])
       {
         t=n[j];
         n[j]=n[j+1];
         n[j+1]=t;
        }
    }
   }
   for(i=2;i<m;i++)
    {
     if(n[i-1]!=n[i]-1)
       {
          printf("%d",n[i]-1);
          break;
       }
  }
 return(0);
 }

How can I do the same without using arrays?

Answer 1

The logic is simple. You just find the sum of the continuous numbers in series for a given n.

And, now add all those numbers provided in the question to find the sum exactly of the given numbers.

The difference is what you can say that difference between those 2 sums is the missing number.

Ex :- Let's say, n = 6.

So, you just find the sum of n consecutive integers starting from 1,...,6 is :- 6 * (6+1) / 2 = 21. Formula of sum of n consecutive integers starting from 1 is {n * (n+1)} / 2.

And, now find the sum of given n-1 numbers.

Say, numbers given are 1,2,4,5,6. Then their sum = 1 + 2 + 4 + 5 + 6 = 18.

Therefore, the missing number = sum of continuous n numbers - sum of given (n-1) numbers = 3.

Answer 2

Find the sum of the given integers. Subtract it from n(n+1)/2.

Explanation: The sum of the first n integers is n(n+1)/2. Therefore the sum of the given integers + missing integer= n(n+1)/2.

Suppose n=10;

Then, 1+2+3+4+5+6+7+8+9+10=10*(10+1)/2==55

If the given integers are- 1,2,3,4,5,6,7,8,10.

Then answer = 55 - (1+2+3+4+5+6+7+8+10)=9.

Answer 3

To find the missing number in the array of 1 to n-1 then we have two methods one is sum formula and second one is use XOR method where both gives O(n) time complexity.

Use Sum Formula

1 Get the sum of numbers

   total = n*(n+1)/2

2 Subtract all the numbers from sum and you will get the missing number.

Use XOR Method

1 XOR all the array elements, let the result of XOR be X1.

2 XOR all numbers from 1 to n, let XOR be X2.

3 XOR of X1 and X2 gives the missing number.

Answer 4

'你可以使用sum方法作为数组中总数的总和，如总数组大小= 5，元素是1,2,4,5,6然后大小= 5然后使用此方法（n（n + 1））/ 2这里n = 5所以15通过计算再次得到数组1 + 2 + 4 + 5 + 6 = 18的元素之和因此18-15 = 3简单'