Python 的 strip() 的运行时间是多少? [英] What's the runtime of Python's strip()?

问题描述

Python 的 strip() 的运行时间是多少?

What's the runtime of Python's strip()?

因为单个字符的remove是O(n)，那么字符串的strip O(n^2)是吗?

Since remove is O(n) for a single char, is strip O(n^2) for a string?

推荐答案

它也只是 O(N).引用与去除空格的纯 strip 对应的代码，来自 2.7.9 版

It is also O(N) only. Quoting the code corresponding to the plain strip which strips the spaces, from the version 2.7.9

Py_LOCAL_INLINE(PyObject *)
do_strip(PyStringObject *self, int striptype)
{
    char *s = PyString_AS_STRING(self);
    Py_ssize_t len = PyString_GET_SIZE(self), i, j;

    i = 0;
    if (striptype != RIGHTSTRIP) {
        while (i < len && isspace(Py_CHARMASK(s[i]))) {
            i++;
        }
    }

    j = len;
    if (striptype != LEFTSTRIP) {
        do {
            j--;
        } while (j >= i && isspace(Py_CHARMASK(s[j])));
        j++;
    }

    if (i == 0 && j == len && PyString_CheckExact(self)) {
        Py_INCREF(self);
        return (PyObject*)self;
    }
    else
        return PyString_FromStringAndSize(s+i, j-i);
}

它首先从左边开始递增变量i，直到找到一个非空格字符，然后从右边开始递减j直到找到一个非空格字符.最后，i 和 j 之间的字符串与 this

It first starts from the left and increments the variable i, till it finds a non-space character and then it starts from the right and decrements j till it finds a non-space character. And finally the string between i and j is returned with this

PyString_FromStringAndSize(s+i, j-i)

但另一方面，strip 删除字符集，有点复杂，但非常相似.

But on the other hand, the strip which removes the set of characters, is slightly complicated but fairly similar.

Py_LOCAL_INLINE(PyObject *)
do_xstrip(PyStringObject *self, int striptype, PyObject *sepobj)
{
    char *s = PyString_AS_STRING(self);
    Py_ssize_t len = PyString_GET_SIZE(self);
    char *sep = PyString_AS_STRING(sepobj);
    Py_ssize_t seplen = PyString_GET_SIZE(sepobj);
    Py_ssize_t i, j;

    i = 0;
    if (striptype != RIGHTSTRIP) {
        while (i < len && memchr(sep, Py_CHARMASK(s[i]), seplen)) {
            i++;
        }
    }

    j = len;
    if (striptype != LEFTSTRIP) {
        do {
            j--;
        } while (j >= i && memchr(sep, Py_CHARMASK(s[j]), seplen));
        j++;
    }

    if (i == 0 && j == len && PyString_CheckExact(self)) {
        Py_INCREF(self);
        return (PyObject*)self;
    }
    else
        return PyString_FromStringAndSize(s+i, j-i);
}

它与前一个相同，但每次都有额外的 memchr(sep, Py_CHARMASK(s[j]), seplen) 检查.所以，这的时间复杂度变成了 O(N * M)，其中 M 是要剥离的实际字符串的长度.

It is the same as the previous one, but it has the extra memchr(sep, Py_CHARMASK(s[j]), seplen) check every time. So, the time complexity of this becomes O(N * M), where M is the length of the actual string of characters to be stripped.

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