0 sub-arrays

Title Description: For array A of length n, the sub-array of the Subarray and close to 0, requires time complexity O (NLOGN).

  1#include <iostream>2#include <algorithm>3#include <math.h>4   5 using namespacestd; 6   7 intFindminsubarry (Const intA[],intN)8 {  9      int*sum =New int[n];Ten     //The first n items of array A and, coexist in the sum array  Onesum[0] = a[0];  A      for(inti =1; I < n; ++i) -     {  -Sum[i] = sum[i-1] +A[i]; the     }  -     intMin1 = Fabs (sum[0]);  -     //The minimum of the absolute value of the first n items of the array a min1  -      for(inti =1; I < n; ++i) +     {  -         if(Fabs (Sum[i]) <min1) +         {  AMin1 =fabs (Sum[i]); at         }  -     }  -Sort (sum, sum +N);  -     //find the minimum value of the difference between the adjacent elements of the sum array min2  -     intMin2 = Fabs (sum[0]-sum[1]);  -      for(inti =2; I < n; ++i) in     {  -         if(Fabs (Sum[i]-sum[i-1]) <min2) to         {  +Min2 = Fabs (Sum[i]-sum[i-1]);  -         }  the     }
delete [] sum;  *     return(Min1 > Min2?)min2:min1);  $ } Panax Notoginseng   - intMainintargcConst Char*argv[]) the {  +     inta[Ten] = { to, - A, -, -, - -, -, the, - the, - at, -};  A     inta2[5] = {-5,2,2,3,2};  the     intmin = Findminsubarry (A,Ten);  +     intmin2 = Findminsubarry (A2,5);  -cout << min <<Endl; $cout << min2 <<Endl; $     return 0;  -}

Results

The calculation of the sum itself and the calculation of the adjacent element difference are all O (N), the sort of sum is O (NLOGN), and therefore the total time complexity: O (Nlogn).

0 sub-arrays