双指针简洁解题法轻松拿下leetcode 581

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2023-12-30 00:21:46

双指针法：巧解最短无序子数组问题

在解决算法问题时，算法的效率至关重要。双指针法是一种巧妙的算法，它利用两个指针同时遍历数据结构，大大提高了算法的效率。本文将深入探讨双指针法，并以 leetcode 581 问题为例，展示其如何巧妙地解决最短无序子数组问题。

双指针法概览

双指针法是一种算法策略，它利用两个指针同时遍历数据结构。这两个指针可以从两端或从中间开始向相反方向移动。通过这种方式，双指针法可以高效地处理各种问题，例如查找子数组、比较序列和检测回文。

解决 leetcode 581 问题

leetcode 581 问题要求我们找到给定数组中最短的无序连续子数组。我们使用双指针法可以高效地解决这个问题。

首先，我们将两个指针分别指向数组的开头和结尾。然后，我们向内移动这两个指针，直到找到第一个无序元素。一旦找到无序元素，我们将继续移动指针，直到找到最后一个无序元素。最后，我们将计算出无序子数组的长度并将其作为结果返回。

Python 代码实现

def find_shortest_unsorted_continuous_subarray(nums):
    """
    Finds the shortest unsorted continuous subarray in a given array.

    Parameters:
    nums: The input array.

    Returns:
    The length of the shortest unsorted continuous subarray.
    """

    # Initialize the left and right pointers.
    left = 0
    right = len(nums) - 1

    # Find the first unsorted element from the left.
    while left < right and nums[left] <= nums[left + 1]:
        left += 1

    # Find the last unsorted element from the right.
    while right > left and nums[right] >= nums[right - 1]:
        right -= 1

    # If the array is already sorted, return 0.
    if left == right:
        return 0

    # Find the minimum and maximum values in the unsorted subarray.
    min_value = min(nums[left:right + 1])
    max_value = max(nums[left:right + 1])

    # Find the first element on the left that is greater than or equal to the minimum value.
    while left > 0 and nums[left - 1] > min_value:
        left -= 1

    # Find the last element on the right that is less than or equal to the maximum value.
    while right < len(nums) - 1 and nums[right + 1] < max_value:
        right += 1

    # Return the length of the unsorted subarray.
    return right - left + 1


# Test the function.
nums = [2, 6, 4, 8, 10, 9, 15]
result = find_shortest_unsorted_continuous_subarray(nums)
print(result)  # Output: 5