Leetcode Daily Question - "Disappearing Numbers"

How are you uu from CSDN, today, Xiao Yalan has opened a new column, and will write questions on Leetcode in the future, try to write one every day, please supervise me! ! ! Alright, let's enter the world of Leetcode

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Leek

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There are many solutions to this question, and Xiao Yalan will analyze it carefully for you next! ! !

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Solution one:

Sort, search in turn, if the next number is not the previous number + 1, then the previous number + 1 is the number that disappeared! ! ! 

But this method, the time complexity is too high, does not meet the requirements of Likou

Using bubble sort, the time complexity is O(N^2) 

Using qsort (quick sort), the time complexity is O(N*logN)

But this question requires a time complexity of O(N)

Well, this method will not work, so Xiao Yalan will not write the code for this method.

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Solution two:

XOR Same as 0, different as 1

•  First, assign x to 0
• for loop, XOR each element of x and nums array
• Then another for loop, let x and the number from 0 to numsSize XOR
• The final answer is the number that appears only once
• The result of the final XOR is the disappearing number we requested

int missingNumber(int* nums, int numsSize)
{
  int i=0;
  int x=0;
  for(i=0;i<numsSize;i++)
  {
     x^=nums[i];
  }
  for(i=0;i<numsSize+1;i++)
  {
      x^=i;
  }
  return x;
}

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Solution three:

using mathematical formulas

The formula for calculating the sum of the first n terms of the arithmetic sequence is: Sn=n*(n+1)/2

After summing, subtract all the values ​​in the nums array, and the final result is the number that disappears

int missingNumber(int* nums, int numsSize)
{
  size_t i=0;
  int x=numsSize*(1+numsSize)/2;
  for(i=0;i<numsSize;i++)
  {
      x-=nums[i];
  }
  return x;
}

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Alright, this is the end of Xiao Yalan's experience in brushing the questions today, let's continue to work hard! ! !