523. Continuous Subarray Sum-优快云博客

本文介绍了一种算法，用于检查给定数组中是否存在连续子数组，其和为给定整数K的倍数。通过使用前缀和与哈希表，该算法能在O(n)时间内解决问题，特别关注了k为0的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

The length of the array won't exceed 10,000.
You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

这题看起来像Subarray Sum Equals K的follow up,不是刚好和为k,而是为K的倍数。主要思路是对presum取余数，判断是否有余数相同的情况，因为数组中数字非负，对于k不等于0时，肯定中间子数组数字的求和为k的整数。这题要注意的是k为0的情况。另外个Subarray Sum Equals K这题一样，为了解决从头开始的情况，hashmap里要放入（0:-1）这对。

代码：

class Solution(object):
    def checkSubarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        hashmap = {0:-1} 
        presum = 0
        for i in xrange(len(nums)):
            presum += nums[i]
            # k为0时要特殊处理
            if k == 0:
                if presum == 0:
                    remainder = 0
                else:
                    return False
            else:    
                remainder = presum % k
            if remainder in hashmap:
                if i - hashmap[remainder] > 1:
                    return True
            else:    
                hashmap[remainder] = i            
        return False

转载于:https://www.cnblogs.com/sherylwang/p/9745373.html