Lintcode:subarray Sum

return The index of the first number and the index of the last number . Examplegiven[return [0, 2] or [1, 3].

Recommended solution: The idea was based on the prefix sum:iterate through the array and for every element array "i", calculate sum of Eleme NTS Form 0 To I (this can simply is done as sum + = arr "I"). If the current sum had been seen before, then there was a zero sum array, the start and end index is returned.

Use Hashmap:o (N) time, but more memory, big case will MLE

1  Public classSolution {2     /**3      * @paramnums:a List of integers4      * @return: A list of integers includes the index of the first number5 * The index of the last number6      */7      PublicArraylist<integer> Subarraysum (int[] nums) {8         //Write your code here9arraylist<integer> res =NewArraylist<integer>();Ten         if(nums==NULL|| nums.length==0)returnRes; OneHashmap<integer, integer> map =NewHashmap<integer, integer>(); AMap.put (0,-1); -         intsum = 0; -          for(inti=0; i<nums.length; i++) { theSum + =Nums[i]; -             if(!map.containskey (sum)) { - map.put (sum, i); -             } +             Else { -Res.add (Map.get (sum) +1); + Res.add (i); A                 returnRes; at             } -         } -         returnRes; -     } -}

Because the above concise code will be MLE, so (Nlog (n)) The second algorithm, a little more time, but less space

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 A      * @return: A list of integers includes the index of the first number at * The index of the last number -      */ -      PublicArraylist<integer> Subarraysum (int[] nums) { -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                 intStart = Math.min (Sums[i].getindex (), Sums[i+1].getindex ()) +1; +                 intEnd = Math.max (Sums[i].getindex (), sums[i+1].getindex ()); A Res.add (start); the Res.add (end); +                 returnRes; -             } $  $         returnRes; -     } -}

Lintcode:subarray Sum