Given an array of integers nums and an integer k, return the total number of continuous subarrays whose sum equals to k

问题

def subarraySum(self, nums: List[int], k: int) -> int:
    count = 0
    target = k
    self.cal(nums,target,count,k)
    return count
def cal(nums,target, count,k):
    if target == 0:
        count = count+1
        target = k
        return count,target
    if target<0:
        return 
        
    for i in range(len(nums)):
        self.cal(nums[:i]+nums[i+1:],target-nums[i],count,k)

''' here when the target < 0 i want to break the loop and for example if there is array 1,2,3 and my target is 2 i will go to the loop first add 1 next again add 2 which is not possible so i want to break the next checking series and want to start from 2'''

回答1:

You could try to use the prefix idea and defaultdict() to solve this Subarray Sum problem more elegantly and efficiently. It's depending on the prefix sum idea, the code is easy to follow, you could try to run it with different data. In the dc[v] we store all prefix sum, the num. of prev. prefix sum with value v. Then it loops and the array to see if new num. w coming has value that satisfies w-v equal k then we got a new count(pair).

from collections import defaultdict
class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        dc = defaultdict(int)
        sum_ = 0
        dc[0] = 1
        result = 0
        for n in nums:
            sum_ += n
            if sum_ - k in dc:
                result += dc[sum_ - k]
            dc[sum_] += 1
        return result

来源：https://stackoverflow.com/questions/65915612/given-an-array-of-integers-nums-and-an-integer-k-return-the-total-number-of-con