python如何在多次递归找到答案后停止接下去的递归

Question

def foo(a):
    if 得到了结果:
        结束递归
        显示
    if 错误:
        return
    else:
        for each in *****:
            foo(each)

这是我大概的思路,试过用exit()虽然停了但会报错

Answer 1

After displaying, return a special value and then check the return variable of foo. If it is that special value, it is directly based on the return value

Answer 2

For problems that are too vague, you can only analyze them through the code.

Answer 3

Recursion requires exit conditions, which is what you call stopping. .

Generally, a branch judgment is required inside recursion, such as:

def fab(n):
  if n<2:
    return 1
  else
    return fab(n-1)+fab(n-2)

After recursing a certain number of times and reaching the above if condition, the recursion ends.

Answer 4

Just add a return after displaying

Answer 5

需要有退出条件, 比如检查一段数据中是否有unicode字符
def check_unicode(data):
    """
    :param data:
    :return:
    """
    if isinstance(data, dict):
        for k, v in data.iteritems():
            if isinstance(v, dict):
                if check_unicode(v):
                    return True
            elif isinstance(v, list):
                if check_unicode(v):    # 检查为True退出
                    return True
            else:
                if v and isinstance(v, unicode):
                    return True
    elif isinstance(data, list):
        for v in data:
            if isinstance(v, dict):
                if check_unicode(v):
                    return True
            elif isinstance(v, list):
                if check_unicode(v):
                    return True
            else:
                if v and isinstance(v, unicode):
                    return True
    else:
        if data and isinstance(data, unicode):
            return True
    return False