和为k的连续子数组

方法一、暴力法

方法二、前缀和 + HashMap

求解和的方式往往能够分解成sum-k方式，查找此key是否在该dict中，count累加计数

# prefix sum
nums = [1,1,1,1]
k, count = 2, 0
dic = {0:1}  # 和为key的前缀和有value个
prefix_sum = 0
for i in range(len(nums)):
    prefix_sum += nums[i]
    if prefix_sum - k in dic:
        count += dic[prefix_sum-k]
    if prefix_sum in dic:
        dic[prefix_sum] += 1
    else:
        dic[prefix_sum] = 1
count

标签：count,key,nums,sum,dic,prefix,连续,数组
来源： https://www.cnblogs.com/demo-deng/p/16615128.html