The sum of the sub-arrays

Given an array of integers, locate and zero the sub-array. Your code should return the starting and ending positions of the subarray that meet the requirements

Sample Example

Given [ -3, 1, 2, -3, 4], return [0, 2] or [1, 3].

Idea: This problem at the beginning of my idea is to use a two-way propaganda to find the first set of consecutive numbers and 0 of the ends, you will find that the complexity of the algorithm is O (n*n); it should be a very complicated method, and then we see this algorithm, we start from the first, to each element of the first n and find out, When sum (j) =sum (i) (J>i) appears, the sum of items i+1 to J is considered to be 0; this is a very ingenious algorithm. So we set up an array of the first n items and we can get the results once in a sort order. Complexity of O (NLOGN)

1 classElementImplementsComparable<element>{2     intindex;3     intvalue;4      PublicElement (intIintv) {5index =i;6Value =v;7     }8      Public intCompareTo (Element other) {9         return  This. value-Other.value;Ten     } One      Public intGetIndex () { A         returnindex; -     } -      Public intGetValue () { the         returnvalue; -     } - } -  Public classSolution { +     /** -      * @paramnums:a List of integers +      * @return: A list of integers includes the index of the first number A * The index of the last number at      */ -      PublicArraylist<integer> Subarraysum (int[] nums) { -         //Write your code here -arraylist<integer> res =NewArraylist<integer>(); -         if(Nums = =NULL|| Nums.length = = 0)returnRes; -         intLen =nums.length; inElement[] sums =NewElement[len+1];  -sums[0]=NewElement ( -1,0); to         intSum =0; +          for(inti=0;i<len;i++){ -Sum + =Nums[i]; thesums[i+1]=NewElement (i,sum); *         } $ Arrays.sort (sums);Panax Notoginseng          for(inti=0;i<len;i++){ -             if(Sums[i].getvalue () ==sums[i+1].getvalue ()) { the                 intm = Math.min (Sums[i].getindex (), Sums[i+1].getindex ()) +1; +                 intn = Math.max (Sums[i].getindex (), sums[i+1].getindex ()); A Res.add (m); the Res.add (n); +                 returnRes; -             } $         } $         returnRes; -     } -}

The sum of the sub-arrays