实现斐波那契数列,求第N项的值swift
func fbnq(_ n: Int) -> UInt64{
if n < 0 || n > 92 {
return 0;
}
var p: UInt64 = 0, c: UInt64 = 1;
for _ in 0 ..< n {
c = c + p;
p = c - p;
}
return c;
}
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function fb(n){
let p = 0, c = 1;
for(let i = 0; i < n; i++){
c = p + c;
p = c - p;
}
return c;
}
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复杂度为 O(n)ui