交换两个数组值使两个数组之差最小----一道面试题

package aAndb_change_to_abs_min;/* * 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]; * */public class Solution {public static void main(String[] strs) {int[] a = {100,99,98,1,2, 3};int[] b = {1, 2, 3, 4,5,40};Solution.Swap2Balance(a, b, 6);for(int n : a)System.err.print(n+" ");System.out.println();for(int n: b)System.out.print(n+" ");}/* * var a=[100,99,98,1,2, 3]; var b=[1, 2, 3, 4,5,40]; * 假设序列a，b中元素的和为sum_a和sum_b。假设aa和bb分别为序列a，b中的元素， * 则交换aa，bb后序列的和变为sum_a-aa+bb，sum_b+aa-bb；两序列的差为（sum_a-aa+bb）-（sum_b+aa-bb）= * sum_a-sum_b-2*（aa-bb）； * 所以可以扫描序列a，b中的元素，找到使abs（sum_a-sum_b-2*（aa-bb））最小的两个元素进行交换，重复此过程， * 直至两序列的差无法减小。 */public static boolean Swap2Balance(int[] pa, int[] pb, int len) {int suma = 0, sumb = 0;for (int i = 0; i < len; i++) {suma += pa[i];sumb += pb[i];}int diff = suma - sumb;while (diff != 0) {int besti = 0, bestj = 0;int bestChange = 0;for (int i = 0; i < len; ++i) {for (int j = 0; j < len; ++j) {int change = pa[i] - pb[j];// 交换后差为(suma-pa[i]+pb[j])-(sumb+pa[i]-pb[j])=diff-2*changeif (Math.abs(diff - 2 * change) < Math.abs(diff - 2 * bestChange)) {besti = i;bestj = j;bestChange = change;}}}if (bestChange == 0)// 差不能再缩小return false;int temp = pa[besti];pa[besti] = pb[bestj];pb[bestj] = temp;suma = suma - bestChange;sumb = sumb + bestChange;diff = suma - sumb;}return true;}}