面试题 10- I. 斐波那契数列

class Solution:
    def fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a % 1000000007

class Solution {
    public int fib(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
}

class Solution {
public:
    int fib(int n) {
        int a = 0, b = 1;
        for (int i = 0; i < n; ++i) {
            int c = (a + b) % 1000000007;
            a = b;
            b = c;
        }
        return a;
    }
};

/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
    let a = 0,
        b = 1;
    for (let i = 0; i < n; ++i) {
        const c = (a + b) % (1e9 + 7);
        a = b;
        b = c;
    }
    return a;
};

func fib(n int) int {
	a, b := 0, 1
	for i := 0; i < n; i++ {
		a, b = b, (a+b)%1000000007
	}
	return a
}

function fib(n: number): number {
    let a: number = 0,
        b: number = 1;
    for (let i: number = 0; i < n; i++) {
        let c: number = (a + b) % 1000000007;
        [a, b] = [b, c];
    }
    return a;
}

impl Solution {
    pub fn fib(n: i32) -> i32 {
        let mut tup = (0, 1);
        for _ in 0..n {
            tup = (tup.1, (tup.0 + tup.1) % 1000000007);
        }
        return tup.0;
    }
}

public class Solution {
    public int Fib(int n) {
        int a = 0, b = 1, tmp;
        for (int i = 0; i < n; i++) {
            tmp = a;
            a = b;
            b = (tmp + b) % 1000000007;
        }
        return a % 1000000007;
    }
}