python -- 序列和之差的绝对值最小

有两个序列a,b，大小都为n,序列元素的值任意整形数，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]之间的差绝对值最小

#! /usr/bin/env python3import sys#aa = [1782, 101, 21, 10, 1, 3]aa = [100 ,99 ,98 ,1 ,2 ,3]#aa = [93, 91, 90, 82, 81, 74, 74, 74, 74, 68]#bb = [23,113, 453, 121, 50, 1178]bb = [1, 2, 3, 4, 5, 40]#bb = [60, 57, 49, 48, 48, 45, 36, 35, 29, 22]div = sum(aa)-sum(bb)def balance(a, b, lenab):global divif div == 0:return;for i in range(lenab):for j in range(lenab):x = a[i] - b[j]A1 = div - 2*xprint("abs(A1) = %d, abs(A) =%d" % (A1, div))if ( abs(A1) < abs(div) ):#交换后，重头开始a[i],b[j] = b[j],a[i]div,A1 = A1,divprint("========A=%d,A1=%d======switch a[%d] b[%d]========================" % (div, A1, i,j))print("sumA = %d " % sum(a), end='')print(a)print("sumB = %d " % sum(b), end='')print(b)print("sumA - sumB = %d " % (sum(a) - sum(b)))balance(a, b, lenab)if __name__ == "__main__":if(len(aa) == len(bb)):A = sum(aa)B = sum(bb)C = sum(aa)-sum(bb)print("------------------source--------------")print("sumA = %d " % A, end='')print(aa)print("sumB = %d " % B, end='')print(bb)print("sumA - sumB = %d " % C)balance(aa, bb,len(aa))print("------------------balance--------------")A = sum(aa)B = sum(bb)C = sum(aa)-sum(bb)print("sumA = %d " % A, end='')print(aa)print("sumB = %d " % B, end='')print(bb)print("sumA - sumB = %d " % C)

复杂度有点大：n*n*n*n,O(n^4)

如果有好的方法，请留言。