Face questions fifty-one: reverse in the array

 

If the array is greater than the number in front of the latter, it is the reverse of the

Method One: Using insertion sort, you can count the number of reverse exchange is the value of the
two: merge sort, you need to sacrifice space makes little to improve the time complexity

public static int InversePairs(int[] data)
         {
             if(data == null || data.length < 0)
                 return 0;
             int[] copy = new int[data.length];
             for(int i = 0; i < data.length; ++i)
                 copy[i] = data[i];
             return  InversePairsCore(data, copy, 0, data.length - 1);   
         }

    public static int InversePairsCore(int[] Data, int [] Copy, int Start, int End) 
         { 
             IF (== Start End) 
             { 
                 Copy [Start] = Data [Start];
                  return 0 ; 
             } 
             int length = (End - Start) / 2 ;
              int left = InversePairsCore (Copy, Data, Start, Start + length);
              int right = InversePairsCore (Copy, Data, Start + length +. 1 , End);
              // I is initialized to the first half of the last digit of the index 
             int I = Start + length;
             // J is initialized to the last half of the number in the subscript 
             int J = End;
              int indexCopy = End;
              int COUNT = 0 ;
              the while (I> J = Start &&> = Start + length +. 1 ) 
             { 
                 IF (Data [ I]> Data [J]) 
                 { 
                     Copy [indexCopy -] = Data [i-- ]; 
                     COUNT + = J - Start - length; 
                 } 
                 the else 
                 { 
                     Copy [indexCopy -] = Data [J, ]; 
                 } 
             } 
             for(; i >= start; --i)
                 copy[indexCopy--] = data[i];
             for(; j >= start + length + 1; --j)
                 copy[indexCopy--] = data[j];
             return left + right + count;
         }