Divide the array into two parts, so that the two parts are the closest and return the difference between the two parts.

[Problem] divides the array into two parts so that the sum of the two parts is the closest, and returns the difference between the two parts. For example:

Int [] array = {,}, divided into two parts: {, 4}, {7}, the difference is 1.

Reference 1: Section 2.18 of the beauty of programming, but the problem is different. Section 2.18 requires an array with a length of 2n to be divided into two arrays with a length of N, so that the two parts and the nearest.

Reference 2: http://www.tuicool.com/articles/ZF73Af

[Idea] a solution for dynamic planning. Obtain the sum of the array. The problem is converted to: select several elements in the array so that the sum of these elements is <= sum/2, and is a set of elements closest to sum/2.

Open an array: int [] [] f = new int [Length + 1] [sum/2 + 1]

State equation: F [I] [J] = max (F [I-1] [J-array [I] + array [I], F [I-1] [J])

Explanation: F [I] [J] indicates the sum of the I elements in the array <= J, and is the element set closest to J. F [I-1] [J-array [I] represents the sum of the I-1 elements in an array that is closest to J-array [I], so f [I] [J] should be the biggest one in [I-1] [J-array [I] + array [I] and f [I-1] [J. It's a bit like a 0-1 backpack problem.

* ** Creation Time: 8:52:57, January 1, October 7, 2014 Project name: Test ** @ author Cao yanfeng * @ since JDK 1.6.0 _ 21 Description: The array is divided into two parts, returns the difference between the two parts. Note: do not divide the array equally */public class dividearraytest {/*** @ Param ARGs */public static void main (string [] ARGs) {// todo auto-generated method stubint [] array = {1, 0, 1, 7, 2, 4}; system. out. println (getmaxdiff (array ));} /* f [I] [J] indicates the maximum capacity that a backpack with an I-element capacity of J can hold */public static int getmaxdiff (INT [] array) {int sum = getsum (array); int length = array. length; int [] [] f = new int [Length + 1] [sum/2 + 1]; for (INT I = 0; I <length; I ++) {for (Int J = 1; j <= sum/2; j ++) {f [I + 1] [J] = f [I] [J]; if (array [I] <= J & F [I] [J-array [I] + array [I]> F [I] [J]) {f [I + 1] [J] = f [I] [J-array [I] + array [I] ;}} return sum-2 * f [length] [sum/2];} public static int getsum (INT [] array) {int sum = 0; For (INT I = 0; I <array. length; I ++) {sum + = array [I] ;}return sum ;}}

Divide the array into two parts, so that the two parts are the closest and return the difference between the two parts.