2016.5.16--leetcode:rotate array,factorial Trailing Zeroe

Rotate Array

This topic Harvest:

Topic:

Rotate an array of n elements to the right by K steps.

For example, with n = 7 and k = 3, the array is [1,2,3,4,5,6,7] rotated to [5,6,7,1,2,3,4] .

Ideas:

My idea: Create a new array to hold the rotated content, but how to store the contents of the original array in the array is unclear.

Leetcode/discuss ideas: A new array, copy the original array, the contents of the new array is stored in the original array, nums[(i + k)%n] = Numscopy[i]

Idea two: Invert the array reversal first bit reverse (nums,nums+n) [7,6,5, 4,3,2,1]

In reverse reverse (nums,nums+k) [5,6,7, 4,3,2,1]

In reverse reverse (nums+k,nums+n) [5,6,7,4,3,2,1]

Code:

Code Listing 1: Ideas 1 time, space complexity are all (n)

1 classSolution2     {3      Public:4         voidRotateintNums[],intNintK//return value is empty5         {6             if(n = =0) || (k <=0))7             {8                 return;//so the returnd is empty.9             }Ten  One             //Make a copy of Nums Avector<int>numscopy (n); -              for(inti =0; I < n; i++) -             { theNumscopy[i] =Nums[i]; -             } -  -             //Rotate the elements. +              for(inti =0; I < n; i++) -             { +nums[(i + k)%n] =Numscopy[i]; A             } at         } -};

Code Listing 2: Thinking two time complexity is (n), spatial complexity is (1)

　

1 void rotate (intint int k) {2     reverse (nums,nums+n);    // explanation See ideas 3     Reverse (nums,nums+k%n); 4     Reverse (nums+k%n,nums+n); 5 }

Factorial Trailing Zeroe

Topic:

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

Given an integer n, find the number of 0 in n!

Ideas:

My idea: Just start to understand the problem, as a n!

Leetcode/dicuss ideas: The idea of a: 0 of the number, is to find 10 of the number, is to find the number of 2*5, 2 occurrences of the number must be more than 5, so is 5 of the number of decisions. Then ask for 5 numbers in N.

Idea two: Suppose N=100,100/5=20, but 100 is not 20 5, but should 20/5=4,20+4=24 5.

Code Listing 1: Ideas 1

1 classSolution {2  Public:3     intTrailingzeroes (intN) {4         intres=0;5          while(n) {//Why Iterate6N/=5;7res+=N;8         }9         returnRes;Ten     } One};

Code Listing 2: Code 2

class Solution {public:    int trailingzeroes (int  n) {           int0;          for (longlong55)             + = n/ i        ; return count;    }};

Daniel's explanation: Https://leetcode.com/discuss/42624/4-lines-4ms-c-solution-with-explanations

Well, to compute the number of trailing zeros, we need to first think clear on what would generate a trailing 0 ? Obviously, a number multiplied by would has 10 a trailing 0 added to it. So we are need to find out how many's would appear in the expression of the 10 factorial. Since 10 = 2 * 5 and there is a bunch more 2 ' s (each even number would contribute at least one 2 ), we only need to Coun t the number of 5 ' s.

Now let's see what numbers would contribute a 5 . Well, simply the multiples of 5 , like 5, ten, +, ... . So is the result simply N/5 ? Well, not. Notice that some numbers could contribute more than one 5 , like 5 * 5 . Well, what numbers would contribute more than one 5 ? Ok, notice that is only multiples of the power of 5 would contribute more than one 5 . For example, multiples of would contribute at least, 5 ' s.

Well, what to count them all? If You try some examples, you may finally get the result, which is n / 5 + n / 25 + n / 125 + ... . The idea behind this expression Is:all 5 the multiples of would contribute one 5 , the multiples of would 25 co Ntribute one more and the multiples of'll contribute another one more ... and so on 5 125 5 . Now, we can write down the following code, which are pretty short.

Code to run with main function:

1#include"stdafx.h"2#include"iostream"3 using namespacestd;4 5 classMyClass6 {7  Public:8     intFactorialtrailingzeroes (intN)9     {Ten         intres =0; One         //cout << res << Endl; A          for(inti =5; I < n; I *=5) -         { -             //cout << i << Endl; //Test theRes + = n/i; -             //cout << res << Endl; -         } -         returnRes; +     } - }; +  A class while at { -  Public: -     intFactorialtrailingzeroes (intN) -     { -         intres =0; -          while(n) in         { -n = n/5; toRes + =N; +         } -         returnRes; the     } * }; $    Panax Notoginseng int_tmain (intARGC, _tchar*argv[]) - { the     //MyClass solution; + while solution; A     intNums; the     intm =0; +CIN >>nums; -m = solution. Factorialtrailingzeroes (Nums);//32, 33 lines are reversed, so I can't get into the for loop. $cout << M <<Endl; $System"Pause"); -     return 0; -}

　　

2016.5.16--leetcode:rotate array,factorial Trailing Zeroe