确定集合中是否存在两个其和刚好为X的元素？

《算法导论》第三版 P22，2.3-7练习题

list1 = [1,2,3,4,5,6,7,8,10,19];count = len(list1);stack4 = []stack2 = []def bsf2(list0, start1, end1, start2, end2, a):            flags = [False, False, False, False];    if len(stack4) > 0:        return True    if start1 == end1 and start2 == end2:        if a == list0[start1] + list0[end2]:            stack4.append(end2)            stack4.append(end2)            stack4.append(start1)            stack4.append(start1)            return True        return False    if list0[start1] + list0[start2] > a or list0[end1] + list0[end2] < a:        return False    mid1 = (int)((start1 + end1) /2);    mid2 = (int)((start2 + end2) /2);    flags[0] = bsf2(list0, start1, mid1, start2, mid2, a);    flags[1] = bsf2(list0, start1, mid1, mid2 + 1, end2, a);    flags[2] = bsf2(list0, mid1 + 1, end1, start2, mid2, a);    flags[3] = bsf2(list0, mid1 + 1, end1, mid2 + 1, end2, a);    if flags[0] and flags[1] and flags[2] and flags[3]:        stack4.append(end2)        stack4.append(start2)        stack4.append(end1)        stack4.append(start1)        return True    if flags[0] and flags[1] and flags[2]:        return bsf2(list0, start1, end1, start2, mid2, a)    if flags[0] and flags[1] and flags[3]:        return bsf2(list0, start1, end1, mid2 + 1, end2, a)    if flags[0] and flags[2] and flags[3]:        return bsf2(list0, start1, mid1, start2, end2, a)    if flags[1] and flags[2] and flags[3]:        return bsf2(list0, mid1 + 1, end1, start2, end2, a)    if flags[0] and flags[2]:        return bsf2(list0, start1, mid1, start2, mid2, a)    if flags[0] and flags[3]:        return bsf2(list0, start1, mid1, mid2 + 1, end2, a)    if flags[1] and flags[2]:        return bsf2(list0, mid1 + 1, end1, start2, mid2, a)    if flags[1] and flags[3]:        return bsf2(list0, mid1 + 1, end1, mid2 + 1, end2, a)def rbsf(list0, start, end, a):    if start == end:        return    if list0[start] + list0[start + 1] > a or list0[end] + list0[end - 1] < a:        return    stack2.append(end);    stack2.append(start);    mid = (int)((start + end) / 2);    rbsf(list0, start, mid, a);    rbsf(list0, mid + 1, end, a);    return;rbsf(list1, 0, count - 1, 9);stack2.reverse();if len(stack2) > 0:    print(stack2);else:    print("find no section")while(len(stack2) > 0):    print(stack2[0:2])    start = stack2.pop(0)    end = stack2.pop(0)    mid = (int)((start + end) / 2);    flag = bsf2(list1, start, mid, mid + 1, end, 9)    if flag:        print(stack4)        break    else:        print("Not found")

算法中，主要使用了递归算法。一个递归算法确定两个加数存在于集合的哪段区域，另一个递归算法确定两个加数分别存在于两个分区的哪个子分区内。

该程序可以找出所有可能存在的加数组合，但是我在找到一个加数组合时就跳出了。