Find subarray with given sum

Given an unsorted array of nonnegative integers, find a continous subarray which adds to a given number. Examples:
Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33

Ouptut: Sum found between indexes 2 and 4

Input: arr[] = {1, 4, 0, 0, 3, 10, 5}, sum = 7

Ouptut: Sum found between indexes 1 and 4

Input: arr[] = {1, 4}, sum = 0

Output: No subarray found.

There may be more than one subarrays with sum as the given sum. The following solutions print first such subarray.

思路：用fast和slow两个下标，计算fast和slow之间的subarray的和，如果和大于给定值，那么把slow向前移动一位，如果和小于给定值，那么把fast向前移动一位。

代码：

/* Returns true if the there is a subarray of arr[] with sum equal to 'sum'
otherwise returns false.  Also, prints the result */
int subArraySum(int arr[], int n, int sum)
{
/* Initialize curr_sum as value of first element
and starting point as 0 */
int curr_sum = arr[0], start = 0, i;

/* Add elements one by one to curr_sum and if the curr_sum exceeds the
sum, then remove starting element */
for (i = 1; i <= n; i++)
{
// If curr_sum exceeds the sum, then remove the starting elements
while (curr_sum > sum && start < i-1)
{
curr_sum = curr_sum - arr[start];
start++;
}

// If curr_sum becomes equal to sum, then return true
if (curr_sum == sum)
{
printf ("Sum found between indexes %d and %d", start, i-1);
return 1;
}

// Add this element to curr_sum
if (i < n)
curr_sum = curr_sum + arr[i];
}

// If we reach here, then no subarray
printf("No subarray found");
return 0;
}