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;}