Majority Element&&Factorial Trailing Zeroes

水题两道

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

class Solution {public:    int majorityElement(vector<int> &num) {        sort(num.begin(),num.end());        if(num[0]==num[num.size()/2])        return num[0];        else if(num[num.size()/2]==num[num.size()-1])        return num[num.size()-1];        else return num[num.size()/2];            }};

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

Note: Your solution should be in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

//思路就是求5的个数。。。。。。。。。。。。。

class Solution {public:    int trailingZeroes(int n) {        int count=0;int temp=n/5;while(temp!=0){count+=temp;temp=temp/5;}return count;            }};