斐波拉切三种求法

# 暴力求解
def func1(num: int):
    if num < 2:
        return num
    return func1(num - 1) + func1(num - 2)


# 去重队列


def recurse(li: list, num: int):
    if num == 0:
        return 0
    if num == 1:
        return 1
    if li[num] != 0:
        return li[num]
    li[num] = recurse(li, num - 1) + recurse(li, num - 2)
    return li[num]


def func2(num: int):
    li = [0 for _ in range(num + 1)]
    return recurse(li, num)

# 双指针
def func3(num:int):
    pre1 = 0
    pre2 = 1
    res = 0
    for i in range(0,num-1): # 少循环2次：0，1
        res = pre1 + pre2
        pre1 = pre2
        pre2 = res
    return res


print(func1(3))
print(func2(3))
print(func3(3))

标签：func1,return,int,recurse,波拉,li,求法,num,三种
来源： https://www.cnblogs.com/bigcoolcool/p/16269540.html