斐波那契数列

public int Fibonacci(int n) {
         if(n==0){
            return 0;
         }else if(n==1){
             return 1;
         }else{
             return Fibonacci(n-1)+Fibonacci(n-2);
         }
   }

class Solution {
public  int Fibonacci(int n) {
        if(n == 0)
            return 0;
        if(n == 1)
            return 1;
        int numfn1 = 0, numfn2 = 1;
        int currentnum;
        for(int i=2; i<=n; ++i) {
            currentnum = numfn1+numfn2;
            numfn1 = numfn2;
            numfn2 = currentnum;
        }
        return currentnum;
    }
};

public class Solution {
    public int Fibonacci(int n) {
        //方法1：用递归，系统会让一个超大的n来让Stack Overflow，所以
        //递归就不考虑了
         
        //使用迭代法，用fn1和fn2保存计算过程中的结果，并复用起来
        int fn1 = 1;
        int fn2 = 1;
         
        //考虑出错情况
        if (n <= 0) {
            return 0;
        }
        //第一和第二个数直接返回
        if (n == 1 || n == 2) {
            return 1;
        }
 
        //当n>=3时，走这里，用迭代法算出结果
        //这里也说明了，要用三个数操作的情况，其实也可以简化为两
        //个数，从而节省内存空间
        while (n-- > 2) {
            fn1 += fn2;
            fn2 = fn1 - fn2;
        }
        return fn1;
    }
}