add 412. Fizz Buzz 🍺

solutions/412. Fizz Buzz.md

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+# [412. Fizz Buzz](https://leetcode.com/problems/fizz-buzz/)
+
+# 思路
+
+这题很简单，我们可以用for循环从i=1到i=n，然后判断i是否是3或者5的倍数。
+
+评论区说因为取余操作`%`比较费时，所以我们应该避免取余，下面代码给出了两种避免取余的思路，亲测确实比直接取余要快一点点。
+
+# C++
+``` C++
+class Solution {
+public:
+    vector<string> fizzBuzz(int n) {
+        vector<string>res(n);
+        for(int i = 1; i <= n; i++) res[i-1] = to_string(i);
+        for(int i = 2; i < n; i+=3) res[i] = "Fizz";
+        for(int i = 4; i < n; i+=5) res[i] = "Buzz";
+        for(int i = 14; i < n; i+=15) res[i] = "FizzBuzz";
+
+        return res;
+    }
+};
+```
+
+``` C++
+class Solution {
+public:
+    vector<string> fizzBuzz(int n) {
+        vector<string>res(n);
+        
+        int fizz = 3, buzz = 5;
+        for(int i = 1; i <= n; i++){
+            if(i == fizz){
+                res[i-1] = "Fizz";
+                fizz += 3;
+                if(i == buzz){
+                    buzz += 5;
+                    res[i-1] += "Buzz";
+                }
+            }
+            else if(i == buzz){
+                    buzz += 5;
+                    res[i-1] = "Buzz";
+            }
+            else res[i-1] = to_string(i);
+        }
+
+        return res;
+    }
+};
+```