大小为n的排序数组

Question

if an array of size n has only 3 values 0 ,1 and 2 (repeated any number of times) what is the best way to sort them. best indicates complexity. consider space and time complexity both

Answer 1

计算每个数字的出现次数，然后用正确的计数填充数组，这是O(n)

Answer 2

Sound's a lot like Dijkstra's Dutch National Flag Problem.

If you need a solution that doesn't use counting, check out http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Sort/Flag/

Answer 3

Untested C-style solution:

int count[3] = {0, 0, 0};

for (int i = 0; i < n; ++i)
    ++count[array[i]];

int pos = 0;
for (int c = 0; c < 3; ++c)
    for (int i = array[c]; i; --i)
        array[pos++] = c;

Answer 4

Haskell two-liner:

count xs = Map.toList $ Map.fromListWith (+) [ (x, 1) | x <- xs ]

countingSort xs = concatMap ( \(x, n) -> replicate n x) (count xs)

> countingSort [2,0,2,1,2,2,1,0,2,1]
[0,0,1,1,1,2,2,2,2,2]