"Sword refers to offer": 1 + 2 + 3 +... + n, "Sword refers to offer"

"Sword refers to offer": 1 + 2 + 3 +... + n, "Sword refers to offer"

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Question link: http://www.nowcoder.com/practice/7a0da8fc483247ff8800059e12d7caf1? Rp = 3 & ru =/ta/coding-interviews & qru =/ta/coding-interviews/question-ranking

Description
Evaluate 1 + 2 + 3 +... + n. It is required that the keyword and condition judgment statement (? B: C ).

Ideas
At the beginning of this question, it seems that we can't think of any great method, but if we are familiar with the features of C ++ or have a better understanding of recursion, we can still find some solutions.

1. Use Recursion

class Solution{public:int Sum_Solution(int n){int sum = n;bool flag = (n>0) && ((sum+=Sum_Solution(n-1))>0);return sum;}};

2. Constructor

class Temp{    public:        Temp()        {            ++N;            Sum+=N;        }        static void Reset()        {            N = Sum = 0;        }        static int GetSum()        {            return Sum;        }    private:        static int N;        static int Sum;}; int Temp::N = 0;int Temp::Sum = 0; class Solution{    public:        int Sum_Solution(int n)        {            Temp::Reset();            Temp *a = new Temp[n];            delete []a;            a = NULL;            return Temp::GetSum();        }};

3. Virtual Functions

class A;A* Array[2]; class A{    public:        virtual int Sum(int n)        {            return 0;        }}; class B:public A{    public:        virtual int Sum(int n)        {            return Array[!!n]->Sum(n-1)+n;        }}; class Solution{    public:        int Sum_Solution(int n)        {            A a;            B b;            Array[0] = &a;            Array[1] = &b;            return Array[1]->Sum(n);        }};

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