Problem

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1: Input: nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

Solution

我們用一個HashMap<preSum,count>來儲存：和為preSum的subarray的數量。 通過遍歷整個輸入的nums陣列，不斷更新HashMap。 在遍歷到時間t時，$nums[0]+nums[1]+...+nums[t-1] = preSum$,在這個時刻之前共有$HashMap[V]$個indices $j$ 使得$nums[0]+nums[1]+...+nums[j] = V$

class Solution:
    def subarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        count = collections.Counter()
        count[0] = 1
        preSum = 0
        ans = 0
        for x in nums:
            preSum += x
            ans += count[preSum-k]
            count[preSum] += 1
        return ans
 

另一道題解法和本題相同：

Problem

In an array A of 0s and 1s, how many non-empty subarrays have sum S?

Example 1:

Input: A = [1,0,1,0,1], S = 2 Output: 4 Explanation: The 4 subarrays are bolded below: [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1]

Note:

A.length <= 30000 0 <= S <= A.length A[i] is either 0 or 1.