LeetCode数组题解：5大经典Python实战
以下是针对LeetCode热门题目Top 100 Liked Questions中“普通数组”类题目的Python版本解法指南。普通数组题目通常涉及数组的基本操作，如遍历、排序、查找等。我将逐步介绍几个代表性题目，提供Python代码和简要解释，帮助您理解和实现。
1. Two Sum（两数之和）
题目描述：给定一个整数数组nums和一个目标值target，在数组中找出和为目标值的两个整数，并返回它们的下标。假设只有一个有效解，且不能使用同一个元素两次。
Python解法：
def twoSum(nums, target):
num_map = {}
for i, num in enumerate(nums):
complement = target - num
if complement in num_map:
return [num_map[complement], i]
num_map[num] = i
return []
代码解释：
- 使用哈希表
num_map存储每个元素的值和下标。 - 遍历数组，计算当前元素的补数（
target - num）。 - 如果补数在哈希表中，直接返回下标对。
- 时间复杂度：$O(n)$，空间复杂度：$O(n)$。
2. Maximum Subarray（最大子数组和）
题目描述：给定一个整数数组nums，找出连续子数组的最大和。
Python解法：
def maxSubArray(nums):
max_sum = float('-inf')
current_sum = 0
for num in nums:
current_sum = max(num, current_sum + num)
max_sum = max(max_sum, current_sum)
return max_sum
代码解释：
- 初始化
max_sum为负无穷和current_sum为0。 - 遍历数组，更新
current_sum为当前元素或当前元素加上之前的子数组和的最大值。 - 同时更新
max_sum为当前最大值。 - 时间复杂度：$O(n)$，空间复杂度：$O(1)$。
3. Move Zeroes（移动零）
题目描述：给定一个数组nums，将所有0移动到数组末尾，同时保持非零元素的相对顺序。
Python解法：
def moveZeroes(nums):
non_zero_idx = 0
for i in range(len(nums)):
if nums[i] != 0:
nums[non_zero_idx], nums[i] = nums[i], nums[non_zero_idx]
non_zero_idx += 1
代码解释：
- 使用双指针：
non_zero_idx指向非零元素应放置的位置。 - 遍历数组，当遇到非零元素时，将其与
non_zero_idx位置的元素交换，并移动指针。 - 时间复杂度：$O(n)$，空间复杂度：$O(1)$。
4. Container With Most Water（盛最多水的容器）
题目描述：给定一个整数数组height，表示垂直线的高度，找出两条线与x轴形成的容器能容纳的最大水量。
Python解法：
def maxArea(height):
left, right = 0, len(height) - 1
max_area = 0
while left < right:
width = right - left
h = min(height[left], height[right])
max_area = max(max_area, width * h)
if height[left] < height[right]:
left += 1
else:
right -= 1
return max_area
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代码解释：
- 使用双指针从数组两端向中间移动。
- 计算当前指针位置形成的容器面积（宽度乘以最小高度）。
- 更新最大面积，并根据高度移动指针。
- 时间复杂度：$O(n)$，空间复杂度：$O(1)$。
5. Product of Array Except Self（除自身以外数组的乘积）
题目描述：给定一个数组nums，返回一个数组，其中每个元素是除自身外所有元素的乘积。要求不使用除法，并在$O(n)$时间内完成。
Python解法：
def productExceptSelf(nums):
n = len(nums)
left_products = [1] * n
right_products = [1] * n
answer = [1] * n
# 计算左侧乘积
for i in range(1, n):
left_products[i] = left_products[i-1] * nums[i-1]
# 计算右侧乘积
for i in range(n-2, -1, -1):
right_products[i] = right_products[i+1] * nums[i+1]
# 合并结果
for i in range(n):
answer[i] = left_products[i] * right_products[i]
return answer
代码解释：
- 使用三个数组：
left_products存储每个元素左侧的累积乘积，right_products存储右侧累积乘积。 - 先从左向右计算左侧乘积，再从右向左计算右侧乘积。
- 最后合并结果得到每个位置的乘积。
- 时间复杂度：$O(n)$，空间复杂度：$O(n)$。
这些题目覆盖了普通数组操作的核心技巧，如双指针、动态规划等。建议您在LeetCode平台上实际练习这些题目，加深理解。如果您有其他具体题目需求，我可以进一步提供帮助！
