二分查找法 - 递归实现
def binSearch(nums, key, low, high):
mid = (low + high) // 2
if low > high:
return -1
if nums[mid] > key:
return binSearch(nums, key, low, mid-1)
elif nums[mid] < key:
return binSearch(nums, key, mid+1, high)
else:
return mid
nums = [2, 4, 5, 9, 11, 16, 19]
print(binSearch(nums, 17, 0, len(nums)))
约瑟夫环问题
def josephRing(num, start):
result = []
nums = [i for i in range(1, num + 1)]
while len(nums) > 0:
result.append(nums[start%len(nums)-1])
nums.remove(nums[start%len(nums)-1])
nums = nums[start-1:] + nums[0: start-1]
return result
print(josephRing(8, 7))
汉诺塔问题（递归）
def hanoi(n, x, y, z):
"""
汉诺塔问题可以分解成一下三个步骤：
将n-1个盘子从x轴借助y轴移动到z轴上；
将第n个盘子从x轴移动到z轴上；
将n-1个盘子从y轴借助x轴移动到z轴上；
"""
if n == 1:
print(x, ' --> ', z)
else:
hanoi(n-1, x, z, y)
print(x, ' --> ', z)
hanoi(n - 1, y, x, z)
hanoi(3, 'X', 'Y', 'Z')
旋转列表（Python）：列表原地修改，参考Leetcode189. 旋转数组
def rotate(nums, target):
for i in range(3):
"""
必须写成nums[:]，否则无法得到正确的值
"""
nums[:] = [nums[-1]] + nums[: -1]
nums = [i for i in range(1, 8)]
rotate(nums, target = 3)
print(nums)
def rotate(nums, k):
k=k%len(nums) #保证循环次数在0-len(nums)之间
"""
必须写成nums[:],否则无法原地修改list，返回值为传入的参数值
"""
nums[:]=nums[len(nums)-k:]+nums[:len(nums)-k] #切割成两块重新组合
