Q32、（数组、规划）

有两个序列 a,b，大小都为 n,序列元素的值任意整数，无序；

要求：通过交换 a,b 中的元素，使[序列 a 元素的和]与[序列 b 元素的和]之间的差最小。

例如: 

var a=[100,99,98,1,2,3];

var b=[1,2,3,4,5,40];

思路：

假设序列a和b的和分别为sumA和sumB，假设sumA>sumB。

交换a中的x和b中的y，交换之后得到的差为：

sumA-x+y-(sumB-y+x) = sumA-sumB-2(x-y)

如果要求差最小，那么abs(sumA-sumB-2(x-y))要最小，也就是x-y要更接近于0.5*(sumA-sumB)

如果一次遍历之后数组没有交换，说明当前交换任何一组数据都会使差增大，循环终止

public class Q32_Notice {public static int Sum(int[] a){int sum = 0;for(int i = 0;i<a.length;i++)sum += a[i];return sum;}public static void getMinDiffer(int[] a,int[] b){int[] big = {};int[] small = {};int differ = 0,min;int i,j,suma,sumb;int flag = 1;while(flag == 1){flag = 0;int ai = 0;int bj = 0;suma = Sum(a);sumb = Sum(b);if(suma == sumb)break;else if(suma > sumb){big = a;small = b;differ = suma - sumb;}else if(suma < sumb){big = b;small = a;differ = sumb - suma;}min = differ;for(i = 0;i< big.length;i++)for(j = 0;j<small.length;j++){int temp = big[i] - small[j];if(Math.abs(differ-2*temp)<min){min = Math.abs(differ-2*temp);flag = 1;ai = i;bj = j;}}if(flag == 1){int mid = big[ai];big[ai] = small[bj];small[bj] = mid;}}}public static void main(String[] args) {int[] a = {100,99,98,1,2,3};int[] b = {1,2,3,4,5,40};getMinDiffer(a, b);System.out.println(Arrays.toString(a));System.out.println(Arrays.toString(b));System.out.println(Sum(a)-Sum(b));}}