【leetcode】斐波那契数列

 

int fib(int n){
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return 1;
    int pre=1;
    int prepre=1;
    int curr = 0;
    for (int i=3; i <= n; i++)
    {
        curr = (pre + prepre) % 1000000007;
        prepre = pre;
        pre = curr;
    }
    return curr;
}