剑指Offer-Python(6-10)

6、旋转数组的最小数字

class Solution:
    def minNumberInRotateArray(self, rotateArray):
        l = len(rotateArray)
        if l == 0:
            return 0

        for i in rotateArray:
            if rotateArray[0] > i:
                return i
        return rotateArray[0]


s = Solution()
t = [3, 4, 5, 1, 2]
min = s.minNumberInRotateArray(t)
print(min)

 

7、斐波那契数列

用递归也写了，但是好像太耗时，牛客这边系统不给过

# -*- coding:utf-8 -*-
class Solution:
    def Fibonacci(self, n):
        # write code here
        f = [0, 1, 1]
        if n <= 1:
            return f[n]
        if n == 2:
            return f[2]
        else:
            for i in range(3, n+1):
                f.append(f[i - 1] + f[i - 2])
            return f[n]


s = Solution()
print(s.Fibonacci(2))
print(s.Fibonacci(3))
print(s.Fibonacci(4))
print(s.Fibonacci(5))

8、跳台阶

因为只有两种步伐（一次跳一阶或一次跳两阶）。比如跳八个台阶时：当最后一步跳一节是一类跳法（此时前七阶的跳法有跳七个台阶的跳法）；当最后一步跳两阶是另一种跳法（此时前6阶的跳法有跳6个台阶的跳法），因此得到规律，if n>2,f(n) = f(n+1)+f(n-2)

class Solution:
    def jumpFloor(self, number):
        f = [0, 1, 2]
        if number <= 2:
            return f[number]
        else:
            for i in range(3, number+1):
                f.append(f[i - 1] + f[i - 2])
            return f[number]


s = Solution()
print(s.jumpFloor(1))
print(s.jumpFloor(2))
print(s.jumpFloor(3))
print(s.jumpFloor(4))
print(s.jumpFloor(5))

9、变态跳台阶

根据第八题思路，由每步可以跨1-n阶，根据最后一步，可以分为n种类型。f(n) = f(1)+f(2)+....+f(n-2)+f(n-1) ;   f(n) = 2*f(n-1)  

class Solution:
    def jumpFloorII(self, number):
        # write code here
        f = [0,1,2]
        for i in range(3,number+1):
            f.append(2*f[i-1])
        return f[number]

s = Solution()
print(s.jumpFloorII(5))

10、矩形覆盖

 

 

# -*- coding:utf-8 -*-
class Solution:
    def rectCover(self, number):
        # write code here
        f = [0, 1, 2]
        for i in range(3, number + 1):
            f.append(f[i - 1] + f[i - 2])
        return f[number]


s = Solution()
print(s.rectCover(4))