<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return <em>the total number of subarrays whose sum equals to</em> <code>k</code>.</p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1], k = 2
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3], k = 3
<strong>Output:</strong> 2
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>-10<sup>7</sup> &lt;= k &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution {
public:
int subarraySum(vector<int>& nums, int k)
{
	int count = 0;
	unordered_map<int, int> sum_count(nums.size());
	sum_count[0]  = 1;
	int sum = 0;
	for (int i = 0; i < nums.size(); ++i)
	{
		sum += nums[i];
		if(sum_count.find(sum - k) != sum_count.end())
			count += sum_count[sum - k];
        int asum = 0;
        if(sum_count.find(sum) != sum_count.end())
            asum = sum_count[sum];
        sum_count[sum] = asum + 1;
	}
	return count;
}
};
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