为什么我们不能计数排序适用于一般的阵列？适用于、阵列
计数排序被称为线性时间如果我们知道，数组中的所有元素都是由上给定数量的限制。如果我们把一个普通的数组，不能我们只是扫描数组中的线性时间，找到数组中的最大值，然后申请计数排序？
Counting sort is known with linear time if we know that all elements in the array are upper bounded by a given number. If we take a general array, cant we just scan the array in linear time, to find the maximum value in the array and then to apply counting sort?
推荐答案
这是不够的，知道的上限运行计数排序：你需要有足够的内存来容纳所有的柜台
It is not enough to know the upper bound to run a counting sort: you need to have enough memory to fit all the counters.
考虑这样一种情况，当你经过的64位整数数组，并找出最大的因素是2 ^ 60。这意味着两件事情：
Consider a situation when you go through an array of 64-bit integers, and find out that the largest element is 2^60. This would mean two things:
您需要的O（2 ^ 60）内存，以及 这是要带O（2 ^ 60）来完成排序。事实上， O（2 ^ 60）是一样的 O（1）帮助不大在这里，因为常数因子是太大了。这是很经常与伪多项式时间算法的问题。
The fact that O(2^60) is the same as O(1) is of little help here, because the constant factor is simply too large. This is very often a problem with pseudo-polynomial time algorithms.
