python--求參賽兩隊全部可能的比賽組合狀況

朋友遇到一個面試題，讓我幫忙實現，題目以下：

紅隊有A1,B1,C1三名隊員，藍隊有A2,B2,C2三名隊員，每輪比賽各隊出一名隊員參加，一名隊員只能參加一次比賽，假設A1不會和B2打，B1不會和B2和C2打，那麼可能出現的組合狀況是什麼？

 

這個面試題的難點在於如何算出全部可能的組合，考慮擴展性的話，還得考慮兩隊人數不相同的問題，所以有了下面代碼：

# coding: utf-8
import copy


def get_team_group_list(team1, team2):
    if len(team1) == 1 or len(team2) == 1:
        team_group_list = list()
        for user1 in team1:
            for user2 in team2:
                user_group = {
                    "U1": user1,
                    "U2": user2
                }
                user_group_list = [user_group]
                team_group_list.append(user_group_list)
        return team_group_list
    else:
        sub_team1 = team1[1:]
        user1 = team1[0]
        team_group_list = list()
        for user2 in team2:
            sub_team2 = filter(lambda x: x != user2, team2)
            sub_team_group_list = get_team_group_list(sub_team1, sub_team2)
            for user_group_list in sub_team_group_list:
                tmp_user_group_list = copy.deepcopy(user_group_list)
                user_group = {
                    "U1": user1,
                    "U2": user2
                }
                tmp_user_group_list.append(user_group)
                team_group_list.append(tmp_user_group_list)
        return team_group_list


def test():
    team1 = ["A1", "B1", "C1"]
    team2 = ["A2", "B2", "C2"]
    exclude_condition_list = [
        {
            "U1": "A1",
            "U2": "B2"
        },
        {
            "U1": "B1",
            "U2": "B2"
        },
        {
            "U1": "B1",
            "U2": "C2"
        }

    ]
    team_group_list = get_team_group_list(team1, team2)
    for num in range(0, len(team_group_list)):
        print("肯能組合{0}:".format(num))
        print(team_group_list[num])
    for num in range(0, len(team_group_list)):
        is_valid = True
        team_group = team_group_list[num]
        for exclude_condition in exclude_condition_list:
            match_list = filter(lambda user_group:
                                (
                                    user_group["U1"] == exclude_condition["U1"] and
                                    user_group["U2"] == exclude_condition["U2"]
                                ),
                                team_group)
            if len(match_list) > 0:
                is_valid = False
        if is_valid:
            print("組合{0}知足條件:".format(num))
            print(team_group_list[num])


test()

顯示效果爲：

肯能組合0:
[{'U1': 'C1', 'U2': 'C2'}, {'U1': 'B1', 'U2': 'B2'}, {'U1': 'A1', 'U2': 'A2'}]
肯能組合1:
[{'U1': 'C1', 'U2': 'B2'}, {'U1': 'B1', 'U2': 'C2'}, {'U1': 'A1', 'U2': 'A2'}]
肯能組合2:
[{'U1': 'C1', 'U2': 'C2'}, {'U1': 'B1', 'U2': 'A2'}, {'U1': 'A1', 'U2': 'B2'}]
肯能組合3:
[{'U1': 'C1', 'U2': 'A2'}, {'U1': 'B1', 'U2': 'C2'}, {'U1': 'A1', 'U2': 'B2'}]
肯能組合4:
[{'U1': 'C1', 'U2': 'B2'}, {'U1': 'B1', 'U2': 'A2'}, {'U1': 'A1', 'U2': 'C2'}]
肯能組合5:
[{'U1': 'C1', 'U2': 'A2'}, {'U1': 'B1', 'U2': 'B2'}, {'U1': 'A1', 'U2': 'C2'}]
組合4 知足條件:
[{'U1': 'C1', 'U2': 'B2'}, {'U1': 'B1', 'U2': 'A2'}, {'U1': 'A1', 'U2': 'C2'}]

經過遞歸方式來解決

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其實啥都是虛的，大家都是來看妹子的，是否是！！！