前言

工作需要，对应需要优化查找子数组和等于特定值的算法

对应的算法推荐：子数组相关题目

以下算法主要针对Python

1. 暴力求解

双重循环时间复杂度为 O(n²)

def subarrays_with_sum_equal_k(nums, k):result = []n = len(nums)for start in range(n):current_sum = 0for end in range(start, n):current_sum += nums[end]if current_sum == k:result.append(nums[start:end + 1])return result# 示例用法
nums = [10, 2, 2, 5, 4]
k = 4
result = subarrays_with_sum_equal_k(nums, k)print("Subarrays with sum equal to {}:".format(k))
for subarray in result:print(subarray)

截图如下所示：

如果是乘积，注意差异之处：

def subarrays_with_product_equal_k(nums, k):result = []n = len(nums)for start in range(n):prod = 1for end in range(start, n):prod *= nums[end]if prod == k:result.append(nums[start:end+1])elif prod > k:breakreturn result# 示例用法
nums = [10, 2, 2, 5, 4]
k = 20
result = subarrays_with_product_equal_k(nums, k)print("Subarrays with product equal to {}:".format(k))
for subarray in result:print(subarray)

2. 前缀和哈希表

时间复杂度为O(n)

def subarrays_with_sum_equal_k(nums, k):result = []prefix_sum = 0 # 用于计算到当前元素的前缀和prefix_sum_map = {0: [-1]}  # 使用哈希表存储前缀和及其索引列表for i, num in enumerate(nums):prefix_sum += numif (prefix_sum - k) in prefix_sum_map:for start in prefix_sum_map[prefix_sum - k]:result.append(nums[start + 1:i + 1])if prefix_sum in prefix_sum_map:prefix_sum_map[prefix_sum].append(i)else:prefix_sum_map[prefix_sum] = [i]return result# 示例用法
nums = [10, 2, -2, -20, 10]
k = -10
result = subarrays_with_sum_equal_k(nums, k)print("Subarrays with sum equal to {}:".format(k))
for subarray in result:print(subarray)

截图如下：

3. 滑动窗口

不适用于负数还有浮点小数的情况

def subarrays_with_sum_equal_k(nums, k):result = []current_sum = 0 # 存储当前窗口内元素的和start = 0 # 滑动窗口的起始位置for end in range(len(nums)):current_sum += nums[end]while current_sum > k and start <= end:current_sum -= nums[start]start += 1if current_sum == k:result.append(nums[start:end+1])return result# 示例用法
nums = [10, 2, 2, 5, 4]
k = 10
result = subarrays_with_sum_equal_k(nums, k)print("Subarrays with sum equal to {}:".format(k))
for subarray in result:print(subarray)

截图如下：