Java to find an array of consecutive n elements to make it and the largest

Given an array, the number of successive elements in the group is the largest value. If all elements are positive, it is obvious that the entire array is required. If the value of all elements is negative, the maximum value is 0.

This is seen on the programming Zhuji, whose time complexity is reduced from O (N3) to O (n).

Java code

package cn.lifx.test;





public class Maxsum


{


public static void Main (string[] args)


　 {


int[] arr = new int[]{31,-41, 59, 26,-53, 58, 97,-93,-23, 84};





maxsum ms = new Maxsum ();





Ms. Max (arr);


Ms. MAX2 (arr);


Ms. MAX3 (arr);





int max = Ms. MAX4 (arr, 0, arr.length-1);


System.out.println ("Max sum is" + max);





Ms. MAX5 (arr);


　 }





//Method 1: Time Complexity of O (n*n*n)


public void Max (int[] arr)


　 {


int max = 0;


int sum = 0;


int left =-1;


int right =-1;





for (int i=0; i<arr.length; i++)


　 {


for (int j=i; j<arr.length; j + +)


　　 {


sum = 0;





for (int k=i; k<=j; k++)


　　 {


sum = sum + arr[k];


　　 }





if (Sum > Max)


　　 {


left = i;


right = J;


max = sum;


　　 }


　　 }


　 }





if (right > 0)


　 {


System.out.println ("+ Left +" ("+ Arr[left] +") to element "+ Right +" ("+ arr[right] + "), Max sum is" + max ";


　 }


Else


　 {


System.out.println ("Max sum is 0.");


　 }


　 }





//Method 2: Time complexity of O (n*n)


public void Max2 (int[] arr)


　 {


int max = 0;


int sum = 0;


int left =-1;


int right =-1;





for (int i=0; i<arr.length; i++)


　 {


sum = 0;


for (int j=i; j<arr.length; j + +)


　　 {


sum = sum + arr[j];


if (Sum > Max)


　　 {


left = i;


right = J;


max = sum;


　　 }


　　 }


　 }





if (right > 0)


　 {


System.out.println ("+ Left +" ("+ Arr[left] +") to element "+ Right +" ("+ arr[right] + "), Max sum is" + max ";


　 }


Else


　 {


System.out.println ("Max sum is 0.");


　 }


　 }





//Method 3: Time complexity of O (n*n)


public void Max3 (int[] arr)


　 {


int max = 0;


int sum = 0;


int left =-1;


int right =-1;





int[] temp = new int[arr.length+1];





temp[0] = 0;


for (int i=0; i<arr.length; i++)


　 {


temp[i+1] = Temp[i] + arr[i];


　 }





for (int i=0; i<arr.length; i++)


　 {


for (int j=i; j<temp.length; j + +)


　　 {


sum = temp[j]-temp[i];


if (Sum > Max)


　　 {


left = i;


right = j-1;


max = sum;


　　 }


　　 }


　 }





if (right > 0)


　 {


System.out.println ("Max" from element "+ Left +" ("+ Arr[left] +") to element "+ Right +" ("+ arr[right] +"), Max Sum is "+ max);


　 }


Else


　 {


System.out.println ("Max sum is 0.");


　 }


　 }





//Method 4: Time Complexity of O (N*LOGN)


public int Max4 (int[] arr, int left, int right)


　 {


int sum = 0;


int max = 0;


int max1 = 0;


int max2 = 0;


int middle = 0;





if (left > right)


　 {


return 0;


　 }


else if (left = right)


　 {


return (Arr[left] > 0? arr[left]: 0);


　 }





middle = (left + right)/2;





for (int i=middle; i>=left; i--)


　 {


sum = sum + arr[i];


if (Sum > Max1)


　　 {


max1 = sum;


　　 }


　 }





sum=0;


for (int i=middle+1; i<=right; i++)


　 {


sum = sum + arr[i];


if (Sum > Max2)


　　 {


max2 = sum;


　　 }


　 }





max = MAX1+MAX2;


int temp1 = Max4 (arr, left, middle);


int temp2 = Max4 (arr, middle+1, right);





if (Temp1 > Max)


　 {


max = Temp1;


　 }





if (Temp2 > Max)


　 {


max = Temp2;


　 }





return Max;


　 }





//Method 5: Time Complexity of O (n)


public void Max5 (int[] arr)


{


int max1 = 0;


int max2 = 0;


int left =-1;


int right =-1;


int temp = 0;





for (int i=0; i<arr.length; i++)


　 {


temp = (max1+arr[i]);


if (temp > 0)


　　 {


max1 = temp;


　　 }


Else


　　 {


left = i+1;


max1 = 0;


　　 }





if (Max1 > Max2)


　　 {


right = i;


max2 = max1;


　　 }


　 }





if (right > 0)


　 {


System.out.println ("+ Left +" ("+ Arr[left] +") to element "+ Right +" ("+ arr[right] + "), Max sum is" + max2);


　 }


Else


　 {


System.out.println ("Max sum is 0.");


　 }


　 }


}

The output is:

Java code

Max　is　from　element　2(59)　to　element　6(97),　max　sum　is　187
Max　is　from　element　2(59)　to　element　6(97),　max　sum　is　187
Max　is　from　element　2(59)　to　element　6(97),　max　sum　is　187
Max　sum　is　187
Max　is　from　element　2(59)　to　element　6(97),　max　sum　is　187