Title: Given an array of arr, the array is unordered, but each value is a positive number, and a positive k is given. All the elements in the sub-array are added by Arr and the oldest array length of K.

Solution: The optimal solution can be the time complexity of O (n), the additional space complexity of O (1). First, the left and right ends of the subarray are labeled with two positions, left and right, starting at the leftmost (left=0,right=0) of the array. The overall process is as follows:

1. Start with variable left=0,right=0, representing sub-array arr[left......right].

2. The variable sum always represents the sum of the sub-array arr[left......right]. At the beginning, sum=arr[0], or arr[0....0].

3. The variable Len keeps track of the length of the largest subarray in all the sub-arrays that accumulate and are k. At first, len=0.

4. The result of the comparison between Sum and K determines whether to move left or right, as follows:

(1) If sum==k, description arr[left.....right] accumulate and K, if arr[left.....right] length is greater than Len, then with the new Len, because all the values of the array are positive, then all start from the left position, The sub-array at the end of the position after right, accumulating and must be greater than K. So let left add 1, which means we start to look at the sub-array starting at the position after left and make sum=sum-arr[left],sum represent Arr[left+1.....right]

(2) If sum is less than K, the description arr[left.....right] also need to add the value behind right, its sum may reach K, so right+1,sum+=arr[right]. It is important to note that the right+1 is out of bounds.

(3) If sum is greater than K, all the sub-arrays that start at the left position and end in position after right are accumulated and must be greater than k, so we examine the sub-array starting at the position immediately after leaving, and sum-=arr[left],sum at this time representing Arr[left +1....right] of the cumulative sum.

5. If right<arr.length, repeat step 4. Otherwise, return directly to Len, the entire process is complete.

The specific reference code is as follows:

1 Public intGetmaxlength (int[] arr,intk)2 {3 if(arr==NULL|| arr.length==0 | | K<=0)4 return0;5 intLeft=0;6 intRight=0;7 intSum=arr[0];8 intLen=0;9 while(right<arr.length)Ten { One if(sum==k) A { -Len=math.max (len,right-left+1); -sum-=arr[left++]; the}Else if(sum<k) - { -right++; - if(right==arr.length) + Break; -sum+=Arr[right]; + } A Else atsum-=arr[left++]; - } - - returnLen; -}

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The oldest array length in the unsorted array and for the given value