![[LeetCode] 560. Subarray Sum Equals K_Medium [LeetCode] 560. Subarray Sum Equals K_Medium](https://image.shishitao.com:8440/aHR0cHM6Ly9ia3FzaW1nLmlrYWZhbi5jb20vdXBsb2FkL2NoYXRncHQtcy5wbmc%2FIQ%3D%3D.png?!?w=700&webp=1)
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].
这个题目思路是利用cumulative sum, 前面数字之和去计算. 参考Leetcode 原题Solution.
建一个diction, 记录 sum 出现过的次数, 然后每次用sum - target, 如果存在在dction里面, count += d[sum-target], 最后返回count. 如果直到这个思路之后其实不难, 否则还蛮难想的.
class Solution:
def subarraySum(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
""" d, count,s = {0:1}, 0,0 # 最开始有0:1 是因为"" 的sum是0, 所以初始为1
for num in nums:
s += num
if s - k in d:
count += d[s-k]
if s not in d:
d[s] = 1
else:
d[s] += 1 return count