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;
}
};