Commit bc5b546

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weiy

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subarray sum equals K medium

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	`1`	`+"""`
	`2`	`+Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.`
	`3`	`+`
	`4`	`+Example 1:`
	`5`	`+Input:nums = [1,1,1], k = 2`
	`6`	`+Output: 2`
	`7`	`+Note:`
	`8`	`+The length of the array is in range [1, 20,000].`
	`9`	`+The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].`
	`10`	`+`
	`11`	`+`
	`12`	`+思路:`
	`13`	`+1. 直接用了暴力的搜索法,TLE。`
	`14`	`+2. Discuss 里用的哈希 + 保留和的方式。`
	`15`	`+具体的是:`
	`16`	`+ pre 一直累加。`
	`17`	`+ 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。`
	`18`	`+`
	`19`	`+这种方法可行。但暂时没搞明白具体的原理。`
	`20`	`+`
	`21`	`+`
	`22`	`+测试地址:`
	`23`	`+https://leetcode.com/problems/subarray-sum-equals-k/description/`
	`24`	`+`
	`25`	`+"""`
	`26`	`+class Solution(object):`
	`27`	`+ def subarraySum(self, nums, k):`
	`28`	`+ """`
	`29`	`+ :type nums: List[int]`
	`30`	`+ :type k: int`
	`31`	`+ :rtype: int`
	`32`	`+ """`
	`33`	`+ dicts = {0:1}`
	`34`	`+`
	`35`	`+ result = 0`
	`36`	`+`
	`37`	`+ pre_sum = 0`
	`38`	`+`
	`39`	`+ for i in nums:`
	`40`	`+ pre_sum += i`
	`41`	`+`
	`42`	`+ result += dicts.get(pre_sum-k, 0)`
	`43`	`+`
	`44`	`+ dicts[pre_sum] = dicts.get(pre_sum, 0) + 1`
	`45`	`+`
	`46`	`+ return result`
	`47`	`+`

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