回溯法解决N皇后的问题

# 使用回溯法解决N皇后的问题

def mk_lst(size):  # 生成棋盘
    return [[0 for i in range(size)] for j in range(size)]


def is_legal_posi(loca, size):  # 判断是否合法坐标
    x, y = loca[0], loca[1]
    size -= 1
    if 0 <= x <= size and 0 <= y <= size:
        return True
    else:
        return False


def place(loca, size, num, lst):  # 放置皇后棋子,num设置1 表示放置皇后，0表示拿走皇后
    lst = lst[:]
    x, y = loca[0], loca[1]
    r = range(-size, size + 1)
    row = [(_, y) for _ in r if is_legal_posi((_, y), size)]
    col = [(x, _) for _ in r if is_legal_posi((x, _), size)]
    up = [(x + _, y + _) for _ in r if is_legal_posi((x + _, y + _), size)]
    down = [(x - _, y + _) for _ in r if is_legal_posi((x - _, y + _), size)]
    for i in row, col, up, down:
        for j in i:
            lst[j[0]][j[1]] = num


def add_next(loca, size, lst):  # 如果到最后一行， 或者下一行没有位置 就返回False
    x, y = loca[0], loca[1]
    if x + 1 > size - 1:  # 说明到底了
        return False
    l = [(x + 1, _) for _ in range(size) if lst[x + 1][_] != 1]
    if not l:
        return False
    else:
        return l


def solution(size):
    path, count_l, outcome, final = [], [], [], []
    for loca in [(0, i) for i in range(size)]:  # 初始位置
        path.append(loca)
    count_l.append(len(path))
    while path:
        lst = mk_lst(size)
        outcome.append(path.pop())
        count_l[-1] -= 1
        for loca in outcome:
            place(loca, size, 1, lst)
        l = add_next(loca, size, lst)
        if l:
            count_l.append(len(l))
            for i in l:
                path.append(i)
        else:
            if len(outcome) == size:
                final.append(outcome[:])
            if outcome:
                outcome.pop()
        t = 0
        while count_l:
            if count_l[-1] == 0:
                count_l.pop()
                t += 1
            else:
                break
        if t:
            for i in range(t):
                if outcome:
                    outcome.pop()

    print(len(final))
    return final


solution(8)