LiteCode-Fizz Buzz(single if)

Description

Given number n. Print number from 1 to n. But:

• when number is divided by 3, print "fizz".
• when number is divided by 5, print "buzz".
• when number is divided by both 3 and 5, print "fizz buzz".

Example

If n = 15, you should return:

[
  "1", "2", "fizz",
  "4", "buzz", "fizz",
  "7", "8", "fizz",
  "buzz", "11", "fizz",
  "13", "14", "fizz buzz"
]

• 题目模板

  class Solution {
  public:
    /**
     * @param n: An integer
     * @return: A list of strings.
     */
    vector<string> fizzBuzz(int n) {
        // write your code here
    }
  };
  

• 题目大意

  题目很简单，在1到n的数字中，能被3和5整除输出”fizz buzz”，能被3整除输出”fizz”，能被5整除输出”buzz”。那么难点在于怎么只用一个if实现。下面先给出正常的代码，再给出一个if版。

• 大体思路

  正常版

  class Solution {
  public:
    /**
     * @param n: An integer
     * @return: A list of strings.
     */
    vector<string> fizzBuzz(int n) {
        // write your code here
        vector<string> v;
        for(int i=1; i<=n; i++){
          if((i%3==0) && (i%5==0))
            v.push_back("fizz buzz");
          else if(i % 3 == 0)
            v.push_back("fizz");
          else if(i % 5 == 0)
            v.push_back("buzz");
          else
            v.push_back(to_string(i));
        }
        return v;
    }
  };
  

  一个if版

  先说下大体思路，就是先把数字全部填进去，然后再对3的倍数进行修改，最后对5的倍数进行修改的时候判断下是否也是3的倍数就可以了。

  class Solution {
  public:
    /**
     * @param n: An integer
     * @return: A list of strings.
     */
    vector<string> fizzBuzz(int n) {
        // write your code here
        vector<string> v;
        for(int i=1; i<=n; i++)
          v.push_back(to_string(i));
  
        for(int i=1; (3*i)<=n; i++)
          v[3*i-1] = "fizz";
  
        for(int i=1; (5*i)<=n; i++){
          if((5*i)%3 == 0)
            v[5*i-1] += " buzz";
          else
            v[5*i-1] = "buzz";
        }
        return v;
    }
  };
  

• 题目链接 https://www.lintcode.com/problem/fizz-buzz/description