Leetcode: Subarray Sum Equals K\\\Binary Subarrays With Sum

最新推荐文章于 2021-07-28 04:51:46 发布

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本文介绍了一种高效算法，用于计算一个整数数组中所有连续子数组的总数，这些子数组的和等于给定整数K。通过使用HashMap记录前缀和的出现次数，可以在O(n)的时间复杂度内解决该问题。

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Problem

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].

Solution

我们用一个HashMap<preSum,count>来保存：和为preSum的subarray的数量。
通过遍历整个输入的nums数组，不断更新HashMap。
在遍历到时间 $t$ 时， $n u m s [0] + n u m s [1] + . . . + n u m s [t - 1] = p r e S u m$ ,在这个时刻之前共有 $H a s h M a p [V]$ ( $V = p r e s u m - k$ )个indices $j$ 使得 $n u m s [0] + n u m s [1] + . . . + n u m s [j] = V$ ，且 $n u m s [j] + . . . + n u m s [t - 1] = p r e S u m - V$ 。
所以在每个时刻我们都将 $H a s h M a p [p r e S u m - k]$ 累加到我们的结果中。

class Solution:
    def subarraySum(self, nums: 'List[int]', k: 'int') -> 'int':
        res = 0
        hashmap = {0:1}
        presum = 0
        for n in nums:
            presum = n+presum
            key = presum-k
            if key in hashmap.keys():
                res += hashmap[key]
            hashmap[presum] = hashmap.get(presum,0)+1
        return res

另一道题解法和本题相同：

Problem

In an array A of 0s and 1s, how many non-empty subarrays have sum S?

Example 1:

Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation:
The 4 subarrays are bolded below:
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]

Note:

A.length <= 30000
0 <= S <= A.length
A[i] is either 0 or 1.