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1.Algorithm

[412]  Fizz Buzz

Easy    Math    String

Given an integer n, return a string array answer (1-indexed) where:

answer[i] == "FizzBuzz" if i is divisible by 3 and 5. answer[i] == "Fizz" if i is divisible by 3. answer[i] == "Buzz" if i is divisible by 5. answer[i] == i (as a string) if none of the above conditions are true.

Example 1:

Input: n = 3
Output: ["1","2","Fizz"]

Example 2:

Input: n = 5
Output: ["1","2","Fizz","4","Buzz"]

Example 3:

Input: n = 15
Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]

Constraints:

1 <= n <= 104

解答 本题是简单题，考察对数组的运算操作

(1).常规解法, 先初始化数组，再对数组位上的值进行比较，时间复杂度 O(n)
func fizzBuzz(n int) []string {
 retVal := make([]string, n)
 for index := 0; index < n; index++ {
  retVal[index] = fmt.Sprintf("%d", index+1)
 }

 step := 15
 for step <= n {
  retVal[step-1] = "FizzBuzz"
  step += 15
 }

 step = 3
 for step <= n {
  if retVal[step-1] != "FizzBuzz" {
   retVal[step-1] = "Fizz"
  }
  step += 3
 }

 step = 5
 for step <= n {
  if retVal[step-1] != "FizzBuzz" {
   retVal[step-1] = "Buzz"
  }
  step += 5
 }
 return retVal
}
(2).对方法（1）的代码优化，直接比较值进行数组的初始化
func fizzBuzz(n int) []string {
 var retVal []string
 for index := 1; index <= n; index++ {
  if index%3 == 0 && index%5 == 0 {
   retVal = append(retVal, "FizzBuzz")
  } else if index%3 == 0 {
   retVal = append(retVal, "Fizz")
  } else if index%5 == 0 {
   retVal = append(retVal, "Buzz")
  } else {
   retVal = append(retVal, fmt.Sprintf("%d", index))
  }
 }
 return retVal
}

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