给定一个数组，从中找出相加之和等于N的所有元素组合

给定一个数组，从中找出相加之和等于N的所有元素组合。

例如，

A=[1,2,3,-3], N = 3

output：共有3种

1,2

3

1,2,3,-3

 

 

循环解法：

def subarraySum(nums, N):
    ans = 0 
    for start in range(len(nums)):
        sum = 0 
        for end in range(start,len(nums)):
            sum += nums[end] 
            if sum == N:
                ans+=1
    return ans
print(subarraySum([1,2,3], 3))

 

前缀和解法:

def subarraySum(nums, N):
    prefix_sum = 0  # 当前的前缀和
    prefix_map = {0: 1}  # 存储前缀和出现的次数（初始值为0出现1次）
    count = 0  # 子数组个数
    for n in nums:
        prefix_sum += n
        # 检查是否存在一个前缀和，使得当前区间的和为 N
        diff = prefix_sum - N
        if diff in prefix_map:
            count += prefix_map[diff]  # 累加满足条件的区间数量
        # 更新当前前缀和出现的次数
        if prefix_sum not in prefix_map:
            prefix_map[prefix_sum] = 0
        prefix_map[prefix_sum] += 1
    return count
 
print(subarraySum([1,2,3], 3))