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 = 0
        for 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 = 1
        for end in range(start, n):
            prod *= nums[end]
            if prod == k:
                result.append(nums[start:end+1])
            elif prod > k:
                break
    
    return 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 += num
        if (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 += 1
        
        if 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：