实现代码：

public class Solution {
    public int MinSubArrayLen(int s, int[] nums) 
    {
       	int? min = null;
    	
    	for(var i = 0;i < nums.Length; i++){
    		var l = MinLen(nums, i, s);
    		if(l.HasValue && l < min || min == null){
    			min = l;
    		}
    	}
    	
    	return min.HasValue ? min.Value : 0;
    }
    
    private int? MinLen(int[] nums, int start, int target)
    {
    	var s = 0;
    	var startFrom = start;
    	while(start < nums.Length){
    		s += nums[start];
    		if(s >= target){
    			return (start - startFrom + 1);
    		}
    		start ++;
    	}
    	
    	return null;
    }
    
}

public class Solution {
    public int MinSubArrayLen(int s, int[] nums) 
    {
       	var len = nums.Length;
    	var sum = 0;
    	int? min = null;
    	
     	// use two pointers 
        var start = 0;
    	var end = 0;
    	
        while (end < len)
        {
    		// if current sum if smaller , keep moving right pointer forward
            while (sum < s && end < len){
                sum += nums[end];
    			end ++;
    		}
     		
    		// if sum is more than target, keep moving left pointer forward to shrink the window size
            while (sum >= s)
            {
    			// find a smaller window then update size
                if (!min.HasValue || end - start < min){
    				min = end - start;
    			}
    			
    			// unload left most value from sum
                sum -= nums[start];
    			start ++;
            }
    		
    		// now sum less than target again
    		// lets play again with the right pointer if not yet reach the end
        }
    	
        return !min.HasValue ? 0 : min.Value;
    }
    
    
}