Continuous Subarray Sum

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:

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

这道题目是求子串的和是k的倍数，我们可以暴力求解，判断每个子串的和是否模k等于0，这里要注意k=0的情况。暴力求解不美观，我们需要一点技巧来重新求解这道题。当a和b除以c的余数相同时，那么(a-b)就一定可以整除c，这个结论很好证明，在此就不再赘述。首先我们可以生成一个新字典，key是余数，value是位置，再新建一个变量保存累加和。通过遍历一次数组，不断地累加元素，当k不等于0时就把累加和(total)除以k的余数存进字典，如果k=0，就直接存total。存之前判断字典中是否已经存在该余数，如果存在，那么就可以返回True，如果不存在，就按照以上步骤存进字典。代码如下：

class Solution(object):
    def checkSubarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        cursum = dict()
        cursum[0] = -1
        total = 0
        for i in range(len(nums)):
            total += nums[i]
            if k != 0:
                total %= k
            if total in cursum:
                if (i-cursum.get(total)) > 1:
                    return True
            else:
                cursum[total] = i
        return False