LintCode题解之斐波纳契数列

public class Solution {
    
    private Integer[] buff = new Integer[1000];
    
    /*
     * @param n: an integer
     * @return: an ineger f(n)
     */
    public int fibonacci(int n) {
        if(buff[n]!=null) return buff[n];
        else if(n==1) return buff[1] = 0;
        else if(n==2) return buff[2] = 1;
        else return buff[n] = fibonacci(n-1) + fibonacci(n-2);
    }
    
}

public class Solution {
    
    /*
     * @param n: an integer
     * @return: an ineger f(n)
     */
    public int fibonacci(int n) {
        if(n==1) return 0;
        else if(n==2) return 1;
        
        int a=0, b=1, c=a+b;
        n-=2;
        
        while(n-->0){
            c = a + b;
            a = b;
            b = c;
        }
        return c;
    }
    
}