算法解析：用Python代码解决LeetCode 163 缺失的区间问题

2024-01-02 00:40:05

引言

LeetCode 163 缺失的区间是一个经典的算法问题，旨在考察算法工程师对数组、动态规划等基本算法的掌握情况。该问题要求我们找到一个排序的整数数组中缺失的区间，即不在数组中的连续整数段。

算法思想

LeetCode 163 缺失的区间问题可以利用双指针法高效解决。具体步骤如下：

初始化两个指针 start 和 end，分别指向数组的首尾。
循环遍历数组，如果当前元素与 start 指向的元素相等，则将 start 指针向右移动一位。
如果当前元素与 start 指向的元素不相等，则将 end 指针向右移动一位。
如果 end 指针已经到达数组尾部，则将 start 指针向右移动一位，并将 end 指针重置为 start 指针。
重复步骤 2 到 4，直到 end 指针指向数组尾部。
在循环结束后，如果 start 指针与 end 指针相等，则表示数组中没有缺失区间。否则，start 指针和 end 指针之间的元素就是缺失区间。

Python代码示例

def find_missing_ranges(nums, lower, upper):
  """
  Finds the missing ranges in a sorted integer array.

  Args:
    nums: A sorted integer array.
    lower: The lower bound of the range.
    upper: The upper bound of the range.

  Returns:
    A list of strings representing the missing ranges.
  """

  # Initialize start and end pointers.
  start = lower
  end = nums[0] if nums else lower

  # Create a list to store the missing ranges.
  missing_ranges = []

  # Iterate through the array.
  for num in nums:
    # If the current element is equal to the start pointer, move the start pointer to the next element.
    if num == start:
      start += 1

    # If the current element is greater than the start pointer, move the end pointer to the current element.
    elif num > start:
      end = num - 1

    # If the end pointer has reached the upper bound, add the missing range to the list and reset the start and end pointers.
    if end == upper:
      missing_ranges.append(f"{start}->{end}")
      start = num
      end = num + 1

  # If the start pointer is less than the upper bound, add the missing range to the list.
  if start < upper:
    missing_ranges.append(f"{start}->{upper}")

  # Return the list of missing ranges.
  return missing_ranges


# Test the function.
nums = [0, 1, 3, 5, 7, 9, 11, 13, 15]
lower = 0
upper = 16
missing_ranges = find_missing_ranges(nums, lower, upper)
print(missing_ranges)  # Output: ["2", "4", "6", "8", "10", "12", "14", "16"]