記一次 TX_Company 的測試開發一面面試題目 (演算法題：第 77 題-爬樓梯)

進入騰訊會議開始：

1、打招呼，介紹自己【自己的工作內容與技能】
2、直接進入主題：
兩道演算法題：
1、寫一個樹的中序排序演算法。【這道題固定模式，沒啥好玩的，自己去熟悉就可以了，沒有下面例子題目有趣】

2、是Leecode的第70. 爬樓梯面試題目：
假設你正在爬樓梯。需要 n 階你才能到達樓頂。
每次你可以爬 1 或 2 個臺階。你有多少種不同的方法可以爬到樓頂呢？
注意：給定 n 是一個正整數。

我的解題思路：

from itertools import permutations
def step_all(n):
    total = 0
    """
    :param n: 總檯階，每次只走1或2，組合有多少種

    :return: all
    """
    if n == 1:
        return 1
    elif n == 2:
        return 2
    elif n >= 3:
        if n % 2 == 0:
            step_min = int(n/2)  # 最小的step登頂
            for i in range(step_min+1, n):
                len_i = i  # 長度
                count_two = (step_min - (len_i - step_min))
                count_one = len_i - count_two
                list_step = [2 for x in range(count_two)]
                list_step.extend([1 for y in range(count_one)])
                li_per = list(permutations(list_step))
                count = len(set(li_per))
                total += count
            return total+2
        elif n % 2 == 1:
            total = 0
            step_min = int((n-1) / 2) + 1  # 最小的step登頂,step-min為2的個數 3:2  4:1   5:0
            for i in range(step_min+1, n):
                len_i = i  # 長度
                count_two = (step_min - (len_i-(step_min-1)))
                print(count_two)
                count_one = len_i - count_two
                list_step = [2 for x in range(count_two)]
                list_step.extend([1 for y in range(count_one)])
                li_per = list(permutations(list_step))
                count = len(set(li_per))
                print(count)
                total += count
            print(total)
            step_min_list = [2 for z in range(step_min-1)]
            step_min_list.extend([1 for i in range(1)])
            li_per = list(permutations(step_min_list))
            step_min_list_count = len(set(li_per))
            total = total + step_min_list_count + 1
            return total

上述的程式碼在1-15之間的執行是沒問題的，所以便提交到Leecode，發現報錯。
然後便換一種方式了，原理不變，藉助數學組合公式並去重得出結果【核心是：對[1,1,1,2,2,1,1]類似的資料進行去重組合】

def step_all(n):
    def mutil_sum(start_x, end_x=None):
        sum_total = 1
        if end_x:
            print(end_x)
            for i in range(start_x, start_x - end_x, -1):
                sum_total *= i
            return sum_total
        for i in range(1, start_x + 1):
            sum_total *= i
        return sum_total
    if n == 1:
        return 1
    elif n == 2:
        return 2
    elif n >= 3:
        if n % 2 == 0:
            total = 0
            min_len = int((n / 2))
            count_two = min_len
            count_one = 0
            for i in range(min_len, n):
                print("只需%s次爬上樓頂"%(count_two + count_one))
                print("2的個數:%s" % count_two)
                print("1的個數:%s" % count_one)
                if count_two >= count_one:
                    type_sum = mutil_sum((count_one+count_two), count_two) / mutil_sum(count_two)
                    total += type_sum
                    print("只需%s次爬上樓頂的組合數為：%s"%((count_two + count_one), type_sum))
                else:
                    type_sum = mutil_sum((count_one + count_two), count_one) / mutil_sum(count_one)
                    total += type_sum
                    print("只需%s次爬上樓頂的組合數為：%s" % ((count_two + count_one), type_sum))
                count_two -= 1
                count_one += 2
            print("只需%s次爬上樓頂" % (count_two + count_one))
            print("2的個數:%s" % count_two)
            print("1的個數:%s" % count_one)
            if count_two >= count_one:
                print("%s >= %s"%(count_two,  count_one))
                type_sum = (mutil_sum((count_one+count_two), count_two)) / mutil_sum(count_two)
                total += type_sum
                print("只需%s次爬上樓頂的組合數為：%s" % ((count_two + count_one), type_sum))
            else:
                type_sum = (mutil_sum((count_one+count_two), count_two)) / mutil_sum(count_one)
                total += type_sum
                print("只需%s次爬上樓頂的組合數為：%s" % ((count_two + count_one), type_sum))
            return total
        if n % 2 == 1:
            total = 0
            min_len = int(((n - 1) / 2)) + 1
            count_two = min_len - 1
            count_one = 1
            for i in range(min_len, n):
                print("只需%s次爬上樓頂"%(count_two + count_one))
                print("2的個數:%s" % count_two)
                print("1的個數:%s" % count_one)
                if count_two >= count_one:
                    type_sum = (mutil_sum((count_one+count_two), count_two)) / mutil_sum(count_two)
                    total += type_sum
                    print("只需%s次爬上樓頂的組合數為：%s"%((count_two + count_one), type_sum))
                else:
                    type_sum = (mutil_sum((count_one+count_two), count_one)) / mutil_sum(count_one)
                    total += type_sum
                    print("只需%s次爬上樓頂的組合數為：%s" % ((count_two + count_one), type_sum))
                count_two -= 1
                count_one += 2
            print("只需%s次爬上樓頂" % (count_two + count_one))
            print("2的個數:%s" % count_two)
            print("1的個數:%s" % count_one)
            if count_two >= count_one:
                type_sum = (mutil_sum((count_one+count_two), count_two)) / mutil_sum(count_two)
                total += type_sum
                print("只需%s次爬上樓頂的組合數為：%s" % ((count_two + count_one), type_sum))
            else:
                type_sum = (mutil_sum((count_one+count_two), count_one)) / mutil_sum(count_one)
                total += type_sum
                print("只需%s次爬上樓頂的組合數為：%s" % ((count_two + count_one), type_sum))
            return total

執行結果OK，程式碼的執行效率底下，需要進行優化，目前只是能實現功能而已。可惜當時沒時間。沒去發現規律。不過我覺得這種更好玩，比那種直接寫二叉樹啥的固定模式好玩。用例子來描述演算法，這方式值得提倡。