LeetCode172 factorial Trailing Zeroes. LeetCode258 Add Digits. LeetCode268 Missing Number

Math problems

172. Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in N!.

Note:your solution should is in logarithmic time complexity. (easy)

Analysis: to find the number of the last 0 of the factorial of N, that is, to find the number of factors in the n! 5 (more than 2:5), the simple idea is to iterate over, for each number, divided by 5 to find the number of factor 5, but will time out.

Considering that a number n is more than 5 of the number of digits divisible by a few (N/5), and then consider divisible by 25, 125 is divisible by the number of numbers, get the following algorithm:

Code:

1 classSolution {2  Public:3     intTrailingzeroes (intN) {4         intsum =0;5          while(N >0) {6Sum + = (N/5);7N/=5;8         }9         returnsum;Ten     } One};

258. Add Digits

Given a non-negative integer num , repeatedly add all its digits until the result have only one digit.

For example:

Given num = 38 , the process is like: 3 + 8 = 11 , 1 + 1 = 2 . Since have only one 2 digit, return it.

Follow up:
Could do it without any loop/recursion in O (1) runtime?

Analysis:

Considering the

Ab% 9 = (9a + a + b)% 9 = (A + b)% 9;

ABC% 9 = (99a + 9 B + A + B + c)% 9 = (A + B + c)% 9;

So it is only a single digit position to use its MoD 9, considering that the number divisible by 9 should return 9 instead of 0, using the first minus repeated plus one way to deal with.

Code:

1 classSolution {2  Public:3     intAdddigits (intnum) {4         if(num = =0) {5             return 0;6         }7         return(Num-1) %9+1;8     }9};

268. Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n , find the one and that's missing from the array.

For example,
Given nums = [0, 1, 3] return 2 . (Medium)

Analysis:

Using the first sum (the first n and), and then the sum of the result and the array and subtract the method, the difference between the number of

Code:

1 classSolution {2  Public:3     intMissingnumber (vector<int>&nums) {4         intn =nums.size ();5         intSUM1 = n * (n +1) /2;6         intsum2 =0;7          for(inti =0; I < nums.size (); ++i) {8Sum2 + =Nums[i];9         }Ten         returnSUM1-sum2; One     } A};

LeetCode172 factorial Trailing Zeroes. LeetCode258 Add Digits. LeetCode268 Missing Number