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

数学题

172. Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity. （Easy）

分析：求n的阶乘中末位0的个数，也就是求n!中因数5的个数（2比5多），简单思路是遍历一遍，对于每个数，以此除以5求其因数5的个数，但会超时。

考虑到一个数n比他小能被5整除的数的个数是一定的（n / 5），由此再考虑能被25整除，125整除的数的个数，得到如下算法：

代码：

 class Solution {

 public:

     int trailingZeroes(int n) {

         int sum = ;

         while (n > ) {

             sum += (n / );

             n /= ;

         }

         return sum;

     }

 };

258. Add Digits

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

For example:

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

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

分析:

考虑到

ab % 9 = (9a + a + b) % 9 = (a + b) % 9;

abc % 9 = (99a + 9 b + a + b + c) % 9 = (a + b + c) % 9;

所以求到其只有个位数位置即用其mod 9即可，考虑到被9整除的数应该返回9而非0，采用先减一再加一方式处理。

代码：

 class Solution {

 public:

     int addDigits(int num) {

         if (num == ) {

             return ;

         }

         return (num - ) %  + ;

     }

 };

268. Missing Number

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

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

分析：

采用先求和（前n项和），再将求和结果与数组和相减的方法，求得差哪个数

代码：

 class Solution {

 public:

     int missingNumber(vector<int>& nums) {

         int n = nums.size();

         int sum1 = n * (n + ) / ;

         int sum2 = ;

         for (int i = ; i < nums.size(); ++i) {

             sum2 += nums[i];

         }

         return sum1 - sum2;

     }

 };

巴特西

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

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