Python递归调用实现数字累加的代码

Z时代
2024-01-10
分类：综合

我就废话不多说了，直接上代码吧！


def sum_numbers(num):
  # 1.出口
  if num == 1:
    return 1
  # 2.数组累加
  temp = sum_numbers(num - 1)
  return num + temp
result = sum_numbers(3)
print(result)

输出：

6

补充拓展：python递归计数及结束递归

题目：搜索旋转排序数组


class Solution:
  TOTAL = 0
  RUN = True
  def search(self, nums: List[int], target: int) -> int:
    # 将数组一分为二，分别比头尾，尾大于头为有序，剩下的为无序
    i, j = 0, len(nums) - 1
    res = -1
    if nums and self.RUN:
      in_middle = (j + i) // 2
      list1 = nums[:in_middle + 1]
      list2 = nums[in_middle + 1:]
      if nums[in_middle] >= nums[i]:
        res = self.binarySearch(list1, target)
        if res == -1:
          self.TOTAL += in_middle + 1
          self.search(list2, target)
        else:
          self.TOTAL += res
      else:
        res = self.binarySearch(list2, target)
        if res == -1:
          self.search(list1, target)
        else:
          self.TOTAL += in_middle + 1 + res
    if not self.RUN:
      return self.TOTAL
    return res
  def binarySearch(self, nums, target):
    """ 二分查找 """
    i, j = 0, len(nums) - 1
    while i <= j:
      in_middle = (j + i) // 2
      if nums[in_middle] == target:
        # print(nums, TOTAL)
        self.RUN = False
        return in_middle
      elif nums[in_middle] < target:
        i = in_middle + 1
      else:
        j = in_middle - 1
    return -1