The largest and most large sub-arrays of an array

Title: The maximum and the number of sub-arrays of an array, which requires O (n) time complexity.

Because of the O (n) time complexity limit, the O (n^2) method of brute force solution is certainly not. Consider recursively the maximum of the subarray of an array a[n], which can be decomposed to the maximum of the a[i] sub-array, and to a condition between a[n-i-1].

• A[n] The maximum sum of the subarray maximum and equal to the A[i] sub-array;
• A[n] Sub-array max and equals a[n-i-1];
• A[n] Sub-arrays Max and cross A[i] and a[n-i-1];

The time complexity of the recursive implementation is O (NLG (n)). Finally, the implementation of dynamic programming with time complexity of O (n) is considered.

/*** Created by Elvalad on 2014/12/4. * To find the maximum and number of sub-arrays of an array*/Importjava.util.ArrayList;ImportJava.util.Scanner; Public classMaxsubarray {/*violent traversal of all sub-arrays and*/     Public Static intMaxSubArray1 (arraylist<integer>a) {intsum = 0; intMax =Integer.min_value;  for(inti = 0; I < a.size (); i++) {sum= 0;  for(intj = i; J < A.size (); J + +) {sum+=A.get (j); if(Sum >max) {Max=sum; }            }        }        returnMax; }    /*Recursive Implementation*/     Public Static intMaxSubArray2 (Arraylist<integer> A,intLowintHigh ) {        intMax =Integer.min_value; intMax1 =Integer.min_value; intMAX2 =Integer.min_value; intMax3 =Integer.min_value; intMAX4 =Integer.min_value; intsum = 0; if(Low <High ) {            intMID = low + (high-low)/2; /*Maximum sub- array maximum for the left half of the Subarray*/Max1=MaxSubArray2 (A, low, mid); /*Maximum sub- array maximum for the right half of the Subarray*/Max2= MaxSubArray2 (A, Mid + 1, high); /*the maximum sub-array spans the mid element, so the maximum value must be the maximum of mid-left plus the maximum of mid-to-right*/             for(inti = mid; I >= low; i--) {sum+=A.get (i); if(Sum >max3) {Max3=sum; }} sum= 0;  for(inti = mid + 1; I <= high; i++) {sum+=A.get (i); if(Sum >max4) {max4=sum; }            }            returnMath.max (Math.max (Max1, Max2), (Max3 +max4)); } Else {            returnA.get (0); }    }    /*Dynamic Planning*/     Public Static intMaxSubArray3 (arraylist<integer>a) {intMax =Integer.min_value; intsum = 0; /*Each element updates the value of sum, zeroing when sum<0, and updating Max if Sum>max*/         for(inti = 0; I < a.size (); i++) {sum+=A.get (i); if(Sum >max) {Max=sum; } Else if(Sum < 0) {sum= 0; }        }        returnMax; }     Public Static voidMain (string[] args) {Scanner scan=NewScanner (system.in); ArrayList<Integer> array =NewArraylist<integer>();  while(Scan.hasnext ()) {Array.add (Scan.nextint ()); } System.out.println ("The Max Sub array is:" +maxSubArray1 (array)); System.out.println ("The Max Sub array is:" + maxSubArray2 (array, 0, Array.size ()-1)); System.out.println ("The Max Sub array is:" +maxSubArray3 (array)); }}

The largest and most large sub-arrays of an array