剑指 Offer II 010. 和为 k 的子数组

题目

输入:nums = [1,1,1], k = 2
输出: 2
解释: 此题 [1,1] 与 [1,1] 为两种不同的情况

解法

• 前缀和 + 哈希表
• 多开辟一点空间
• preSum[j] = k + preSum[i] ; k = a[i+1] + ···+a[j] =>子数组和

class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        # +- no using window 
        # using prefix sum
        # preprocessing
        n = len(nums)
        preSum = [0]* (n+1)
        pre= 0
        for i in range(n):
            pre += nums[i]
            preSum[i+1] = pre
        
        # main using hash map / dictionary
        dic = {}
        count = 0
        for i in range(n+1):
            if preSum[i] - k in dic:
                count += dic[preSum[i] - k]
            if preSum[i] not in dic:
                dic[preSum[i]] = 1
            else:
                dic[preSum[i]] += 1
        return count