20170603-leetcode-560-Subarray Sum Equals K

1.Description

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].

解讀
給定一個整數數組，和一個整數k，求有多少個連續子序列的和爲k。

2.Solution

解題思路：
最能想到的辦法就是遍歷了，存儲前n項的和，然後遞減回去,但是超時了

class Solution(object):
    def subarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if nums == []: return 0

        sumlist, N = [nums[0]], 0
        for i in range(1, len(nums)):
            sumlist.append(sumlist[-1] + nums[i])
        for i in range(len(sumlist)):
            if sumlist[i] == k:
                N = N + 1
            for j in range(i):
                if sumlist[i] - sumlist[j] == k:
                    N = N + 1
        #print(N)
        return N

那麼可以想到：通過記錄前n項和的次數，來快速得到解，什麼意思呢，
比如我們已經把前面n-1項和的次數放入到了詞典中，比如{S1 :2, S2 :3, S3 :3 … Si :V…Sn−1:k′ }，這時候新來一個Sn , 如果Sn -K(題目中給的值：要求的和)=Si ，恰好在上面次數詞典中，我們就可以直接把次數V提取出來。

class Solution(object):
    def subarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        count,cur,res={0:1},0,0
        for num in nums:
            cur=cur+num#計算前n項和
            # (如果前n項和)cur-(需要求的和)k in (前n項和次數)詞典中，那麼get出來次數
            res=res+count.get(cur-k,0)
            count[cur]=count.get(cur,0)+1 #(如果前n項和)cur(的次數)在字典中加1
        return res