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.
解题思路
本题和Contiguous Array很类似，解题思路基本一致，但一开始要想到使用hash table的思路也不是很容易，本题的核心思想是要想到使用求余的方法，遍历数组，依次叠加当前数组元素并将和对K求余，求余结果存入哈希表中，如果遍历到当前位置，求余结果存在于哈希表中，则表明上一次求余结果相同的位置到当前位置的子数组的和为k的倍数，如果位置相差> 1,则返回true.
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
unordered_map<int,int> hash;
int sum = 0;
hash[0] = -1;
for(int i = 0; i < nums.size(); ++i)
{
sum += nums[i];
if(k != 0)
sum %= k;
if(hash.find(sum) != hash.end())
{
if(i - hash[sum] > 1) return true;
}else
{
hash[sum] = i;
}
}
return false;
}
};
