[算法]:二分法-查找有序数组中一个数字位置

#问题
二分查找

list.index()无法应对大规模数据的查询,需要用其它方法解决,这里谈的就是二分查找

#思路说明

在查找方面,python中有list.index()的方法。例如:

>>> a=[2,4,1,9,3] #list可以是无序,也可以是有序
>>> a.index(4) #找到后返回该值在list中的位置
1
>>> a.index(5) #如果没有该值,则报错
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 5 is not in list

这是python中基本的查找方法,虽然简单,但是,如果由于其时间复杂度为O(n),对于大规模的查询恐怕是不足以胜任的。二分查找就是一种替代方法。

二分查找的对象是:有序数组。这点特别需要注意。要把数组排好序先。怎么排序,可以参看我这里多篇排序问题的文章。

基本步骤:

1. 从数组的中间元素开始,如果中间元素正好是要查找的元素,则搜素过程结束;
2. 如果某一特定元素大于或者小于中间元素,则在数组大于或小于中间元素的那一半中查找,而且跟开始一样从中间元素开始比较。
3. 如果在某一步骤数组为空,则代表找不到。

这种搜索算法每一次比较都使搜索范围缩小一半。时间复杂度:O(logn)
# 递归
def binary_search(lst, value, lo, hi):
    if lo > hi:
        return -1
    half = (lo + hi)/2
    if lst[half] == value:
        return half
    elif lst[half] > value:
        return binary_search(lst, value, lo, half-1)
    else:
        return binary_search(lst, value, half+1, hi)


# 循环
def binary_search_while(lst, value):
    lo, hi = 0, len(lst)-1
    while lo <= hi:
        half = (lo + hi)/2
        if lst[half] > value:
            hi = half - 1
        elif lst[half] < value:
            lo = half + 1
        else:
            return half
    return -1

 


对于python,不能忽视其强大的标准库。经查阅,发现标准库中就有一个模块,名为:bisect。


- 模块接受排序后的列表。
- 本模块同样适用于长列表项。因为它就是用二分查找方法实现的,有兴趣可以看其源码(源码是一个很好的二分查找算法的例子,特别是很好地解决了边界条件极端的问题.)
- 关于bisect模块的更多内容,可以参看[官方文档](https://docs.python.org/2/library/bisect.html)

"""Bisection algorithms."""

def insort_right(a, x, lo=0, hi=None):
    """Insert item x in list a, and keep it sorted assuming a is sorted.

    If x is already in a, insert it to the right of the rightmost x.

    Optional args lo (default 0) and hi (default len(a)) bound the
    slice of a to be searched.
    """

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x < a[mid]: hi = mid
        else: lo = mid+1
    a.insert(lo, x)

insort = insort_right   # backward compatibility

def bisect_right(a, x, lo=0, hi=None):
    """Return the index where to insert item x in list a, assuming a is sorted.

    The return value i is such that all e in a[:i] have e <= x, and all e in
    a[i:] have e > x.  So if x already appears in the list, a.insert(x) will
    insert just after the rightmost x already there.

    Optional args lo (default 0) and hi (default len(a)) bound the
    slice of a to be searched.
    """

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if x < a[mid]: hi = mid
        else: lo = mid+1
    return lo

bisect = bisect_right   # backward compatibility

def insort_left(a, x, lo=0, hi=None):
    """Insert item x in list a, and keep it sorted assuming a is sorted.

    If x is already in a, insert it to the left of the leftmost x.

    Optional args lo (default 0) and hi (default len(a)) bound the
    slice of a to be searched.
    """

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x: lo = mid+1
        else: hi = mid
    a.insert(lo, x)


def bisect_left(a, x, lo=0, hi=None):
    """Return the index where to insert item x in list a, assuming a is sorted.

    The return value i is such that all e in a[:i] have e < x, and all e in
    a[i:] have e >= x.  So if x already appears in the list, a.insert(x) will
    insert just before the leftmost x already there.

    Optional args lo (default 0) and hi (default len(a)) bound the
    slice of a to be searched.
    """

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x: lo = mid+1
        else: hi = mid
    return lo

# Overwrite above definitions with a fast C implementation
try:
    from _bisect import *
except ImportError:
    pass

 参考资料:https://github.com/qiwsir/algorithm

 
posted @ 2017-07-07 14:51  懒惰的咕噜  阅读(2956)  评论(0编辑  收藏  举报