题目描述：

LeetCode 523. 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.

题目大意：

求数组nums中是否存在k的整数倍，并且长度至少为2的连续子段和。

注意：

1. 数组长度不超过10,000。
2. 可以假设所有数字的和范围在32位带符号整数之内。

解题思路：

遍历数组nums，求前i项和total；对k取模，记模值为m

利用dmap[m]记录模为m的前i项和的最小下标，初始令dmap[0] = -1

若dmap[m] + 1 < i，则返回True

Python代码：

class Solution(object):
    def checkSubarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        dmap = {0 : -1}
        total = 0
        for i, n in enumerate(nums):
            total += n
            m = total % k if k else total
            if m not in dmap: dmap[m] = i
            elif dmap[m] + 1 < i: return True
        return False

 

本文链接：http://bookshadow.com/weblog/2017/02/26/leetcode-continuous-subarray-sum/
请尊重作者的劳动成果，转载请注明出处！书影博客保留对文章的所有权利。

如果您喜欢这篇博文，欢迎您捐赠书影博客： ，查看支付宝二维码