斐波那契的小改动。
impl Solution { pub fn tribonacci(n: i32) -> i32 { if n == 0 { return 0; } else if n == 1 { return 1; } else if n == 2 { return 1; }
let mut data = vec![0, 1, 1]; let mut next = 0;
let mut i = 2; while i < n { data[next] = data[0] + data[1] + data[2]; i += 1; if i == n { return data[next]; }
next = (next + 1) % 3; } data[next] }}