int32位 posted @ Mar 22, 2016 04:03:09 PM in algorithm , 2478 阅读




  • 空指针检查(包括src和dest)
  • 内存重叠,要检查指针是否重叠
  • 最后拷贝时,别忘了在dest追加字符串终结符号0
  • 如何保证dest已分配足够内存
  • 为什么从后向前拷贝?


  • 泛型参数(泛化编程)
  • 类型必需可比较或者提供比较器
  • 不能实例泛型类型,即不能new T[]或者new T(),只能使用Object数组
  • 内存动态分配,内存拷贝(数组拷贝)


import random
def sample(filename, n = 3):
    assert(filename is not None)
    result = []
    with open(filename) as f:
        lines = f.readlines()
    while len(result) < n :
        line = random.choice(lines)
        if line not in result:
    return result


  • 必须把文件内容全读入内存,如果文件很大怎么办?
  • 设N为文件行数,n和N接近时,越到后面,冲突越大,效率极低。
  • 有人说,先随机生成n个数构成一个集合A(还是有冲突哦),读取文件,当行数属于A时,取出该行,问题不就解决了? 问题是文件总行数是多少?必须知道文件总行数才能知道随机数的取值范围啊。必须先扫描一遍文件?文件很大时这样的效率如何?



蓄水池抽样(Reservoir Sampling )是一个很有趣的问题,它能够在o(n)时间内对n个数据进行等概率随机抽取, 例如:从1000个数据中等概率随机抽取出100个。另外,如果数据集合的量特别大或者还在增长(相当于未知数据集合总量), 该算法依然可以等概率抽样。



  1. 从文件中取前n行,结果集result
  2. 令i表示当前行数,c为当前行内容,m = random(1, i),即m为1~i的一个随机数。
  3. 若m <= n, 则令result[m] = c,即替换第m行内容,否则舍弃c。
  4. 若已到文件结尾,退出算法,返回result,否则转2

利用归纳法可以证明每行被取的概率是相等的,证明过程 见handspeaker博客



import random
def sample(filename, n = 5):
    assert(filename is not None)
    result = []
    with open(filename, mode = "r") as f:
        for i in range(n):
            line = f.readline()
            if line == '':
                return result
        i = n
        for line in f:
            k = random.randint(0, i)
            if k < n:
                result[k] = line.strip()
            i += 1
    return result


  • 无匹配
  • 无匹配
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