From bc5b5461de059f0a4bfa2534039f4241bbcf5166 Mon Sep 17 00:00:00 2001
From: weiy <b754048538@163.com>
Date: Sat, 27 Oct 2018 17:53:18 +0800
Subject: [PATCH] subarray sum equals K medium

---
 Array/SubarraySumEqualsK.py | 47 +++++++++++++++++++++++++++++++++++++
 1 file changed, 47 insertions(+)
 create mode 100644 Array/SubarraySumEqualsK.py

diff --git a/Array/SubarraySumEqualsK.py b/Array/SubarraySumEqualsK.py
new file mode 100644
index 0000000..2d354b2
--- /dev/null
+++ b/Array/SubarraySumEqualsK.py
@@ -0,0 +1,47 @@
+"""
+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].
+
+
+思路：
+1. 直接用了暴力的搜索法，TLE。
+2. Discuss 里用的哈希 + 保留和的方式。
+具体的是：
+   pre 一直累加。
+   如果 pre - k 存在于保存的字典中，那么结果里加上这个次数即可。
+
+这种方法可行。但暂时没搞明白具体的原理。
+
+
+测试地址：
+https://leetcode.com/problems/subarray-sum-equals-k/description/
+
+"""
+class Solution(object):
+    def subarraySum(self, nums, k):
+        """
+        :type nums: List[int]
+        :type k: int
+        :rtype: int
+        """
+        dicts = {0:1}
+        
+        result = 0
+        
+        pre_sum = 0
+        
+        for i in nums:
+            pre_sum += i
+            
+            result += dicts.get(pre_sum-k, 0)
+            
+            dicts[pre_sum] = dicts.get(pre_sum, 0) + 1
+
+        return result
+        
\ No newline at end of file