【LeetCode523】. Continuous Subarray Sum

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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.
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思想：
1.长度分别为2.3.4...size
2.分别求  如果满足sum%k==0  return true,否则返回false
3.注意处理k==0情况
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bool continuousSubarraySum(vector<int> nums,int k)
{

int size = nums.size();
for (int ii = 2; ii <= size; ++ii)
{
int sum = 0;
for (int jj = 0; jj < ii; ++jj)
{
sum = sum + nums[jj];
}
if (k == 0)
{
if (sum == 0)
return true;
}
else
{
if (sum%k == 0)
return true;
}

for (int kk = ii; kk < size; ++kk)
{
sum = sum - nums[kk - ii] + nums[kk];
if (k == 0)
{
if (sum == 0)
return true;
}
else
{
if (sum%k == 0)
return true;
}

}

}

return false;
}