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 = 111 + 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;
}
};
 

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