代码之家 › 专栏 › 技术社区 › PowerApp101

从有序字符序列递归生成有序子字符串?

sequences substring recursion algorithm c++

3

PowerApp101 · 技术社区 · 17 年前

得到答案后编辑

假设我有一个由字符s[0]:s[N]组成的字符串s,其中每个字符s[I]<=s[i+1]

aaacdddghzz

例如,我会

a
aa
aaa
ad
aad
aaad
add
aadd
aaadd
addd
aaddd
aaaddd
d
dd
ddd
.
.
.
ac
aac
.
.
.
acdddghzz
aacdddghzz
aaacdddghzz

但不是

ca
hdz
...etc

现在我知道如何计算出有多少个组合。创建字符串中字母频率的直方图。所以在上面的例子中

a=3
d=3
c=1
g=1
h=1
z=2

公式是 (a+1)(c+1)(d+1)(g+1)(h+1)(z+1) = 4*4*2*2*2*3 = 384 . 有384个子字符串保持s[i]<=s[i+1]关系。

5 回复 | 直到 17 年前

1

5

Salman A 17 年前

以上是Ryan Shaw的回答:

根据每个字母的数量,以基数计算每个数字,而不是以二进制进行计数。例如:

a d c g h z
3 3 1 1 1 2

因此,我认为:

0 0 0 0 0 0
0 0 0 0 0 1
0 0 0 0 0 2
0 0 0 0 1 0
0 0 0 0 1 1 
0 0 0 0 1 2
0 0 0 1 0 0
...
0 0 0 1 1 2
0 0 1 0 0 0
...
0 0 1 1 1 2
0 1 0 0 0 0 
...
0 3 1 1 1 2
1 0 0 0 0 0
...
3 3 1 1 1 2

1 2 0 0 1 1 => addhz
3 0 0 0 1 2 => aaahzz

以及守则:

void GetCounts(const string &source, vector<char> &characters, vector<int> &counts)
{
    characters.clear();
    counts.clear();

    char currentChar = 0;
    for (string::const_iterator iSource = source.begin(); iSource != source.end(); ++iSource)
    {
        if (*iSource == currentChar)
            counts.back()++;
        else
        {
            characters.push_back(*iSource);
            counts.push_back(1);
            currentChar = *iSource;
        }
    }
}

bool Advance(vector<int> &current, const vector<int> &max)
{
    if (current.size() == 0)
        return false;

    current[0]++;
    for (size_t index = 0; index < current.size() - 1 && current[index] > max[index]; ++index)
    {
        current[index] = 0;
        current[index + 1]++;
    }
    if (current.back() > max.back())
        return false;
    return true;
}

string ToString(const vector<int> &current, const vector<char> &characters)
{
    string result;
    for (size_t index = 0; index < characters.size(); ++index)
        for (int i = 0; i < current[index]; ++i)
            result += characters[index];
    return result;
}

int main() { 
    vector<int> max;
    vector<char> characters;

    GetCounts("aaadddcghzz", characters, max);

    vector<int> current(characters.size(), 0);
    int index = 1;
    while (Advance(current, max))
    {
        cout << index++ << ":" << ToString(current, characters) << endl;
    }
}

2

Dave 17 年前

下面是生成所有子序列的递归算法。

/* in C -- I hope it will be intelligible */

#include <stdio.h>

static char input[] = "aaabbbccc";
static char output[sizeof input];

/* i is the current index in the input string
 * j is the current index in the output string
 */
static void printsubs(int i, int j) {
    /* print the current output string */
    output[j] = '\0';
    printf("%s\n", output);
    /* extend the output by each character from each remaining group and call ourselves recursively */
    while(input[i] != '\0') {
        output[j] = input[i];
        printsubs(i + 1, j + 1);
        /* find the next group of characters */
        do ++i;
        while(input[i] == input[i - 1]);
    }
}

int main(void) {
    printsubs(0, 0);
    return 0;
}

如果您的兴趣仅仅是计算有多少子序列,那么您可以更有效地进行计算。只需计算每个字母的数量,每个值加1,然后将它们相乘。在上述示例中,对于(3+1)*(3+1)*(3+1)*(3+1)*(2+1)=192个子序列,有3个a、3个b、3个c和2个d。这样做的原因是,您可以在0和3A、0和3B、0和3C、0和2D之间进行选择,以及 所有这些选择都是独立的

3

1

yinyueyouge 17 年前

考虑到集合{a,a,a,d,d,d,c,g,h,z,z},您的目标是按顺序列出其所有唯一子集,除了空集: {a} {a,a,a} {a,a,a,d}

{}     = 000
{C}    = 001
{B}    = 010
{BC}   = 011
{A}    = 100
{AC}   = 101
{AB}   = 110
{ABC}  = 111

看到模式了吗?只需使用一个从0增长到2^n-1的整数。如果整数的第i位是1,则从集合中提取第i个元素。

注意:因为在您的示例中,字符串中有重复项;因此,生成后,可能需要删除重复项。

4

0

cjs 17 年前

嗯,在我看来,有一种解决方案与您的类似,但与您的输出不匹配(请参见我对问题的评论),就是简单地遍历原始字符串的尾部列表(例如,对于“abc”,遍历“abc”、“bc”和“c”),并为每个字符串生成前缀列表(“abc”、“ab”、“a”,然后是“bc”、“b”,然后是“c”)。这和你想要的相比如何?

5

0

Ray Tayek 17 年前

我使用了这个java代码( http://www.merriampark.com/comb.htm )结果只有383个。代码生成了太多的副本,所以我不得不扔掉很多。我只得到了383分(见下文)。您可能想看看STL中的下一个组合的C++代码(但是我很难在任何地方找到源代码)。电源组可能是最好的方法(但也可能有重复的方法)。

a
aa
aaa
aaac
aaacg
aaacgh
aaacghz
aaacghzz
aaacgz
aaacgzz
aaach
aaachz
aaachzz
aaacz
aaaczz
aaad
aaadc
aaadcg
aaadcgh
aaadcghz
aaadcghzz
aaadcgz
aaadcgzz
aaadch
aaadchz
aaadchzz
aaadcz
aaadczz
aaadd
aaaddc
aaaddcg
aaaddcgh
aaaddcghz
aaaddcghzz
aaaddcgz
aaaddcgzz
aaaddch
aaaddchz
aaaddchzz
aaaddcz
aaaddczz
aaaddd
aaadddc
aaadddcg
aaadddcgh
aaadddcghz
aaadddcghzz
aaadddcgz
aaadddcgzz
aaadddch
aaadddchz
aaadddchzz
aaadddcz
aaadddczz
aaadddg
aaadddgh
aaadddghz
aaadddghzz
aaadddgz
aaadddgzz
aaadddh
aaadddhz
aaadddhzz
aaadddz
aaadddzz
aaaddg
aaaddgh
aaaddghz
aaaddghzz
aaaddgz
aaaddgzz
aaaddh
aaaddhz
aaaddhzz
aaaddz
aaaddzz
aaadg
aaadgh
aaadghz
aaadghzz
aaadgz
aaadgzz
aaadh
aaadhz
aaadhzz
aaadz
aaadzz
aaag
aaagh
aaaghz
aaaghzz
aaagz
aaagzz
aaah
aaahz
aaahzz
aaaz
aaazz
aac
aacg
aacgh
aacghz
aacghzz
aacgz
aacgzz
aach
aachz
aachzz
aacz
aaczz
aad
aadc
aadcg
aadcgh
aadcghz
aadcghzz
aadcgz
aadcgzz
aadch
aadchz
aadchzz
aadcz
aadczz
aadd
aaddc
aaddcg
aaddcgh
aaddcghz
aaddcghzz
aaddcgz
aaddcgzz
aaddch
aaddchz
aaddchzz
aaddcz
aaddczz
aaddd
aadddc
aadddcg
aadddcgh
aadddcghz
aadddcghzz
aadddcgz
aadddcgzz
aadddch
aadddchz
aadddchzz
aadddcz
aadddczz
aadddg
aadddgh
aadddghz
aadddghzz
aadddgz
aadddgzz
aadddh
aadddhz
aadddhzz
aadddz
aadddzz
aaddg
aaddgh
aaddghz
aaddghzz
aaddgz
aaddgzz
aaddh
aaddhz
aaddhzz
aaddz
aaddzz
aadg
aadgh
aadghz
aadghzz
aadgz
aadgzz
aadh
aadhz
aadhzz
aadz
aadzz
aag
aagh
aaghz
aaghzz
aagz
aagzz
aah
aahz
aahzz
aaz
aazz
ac
acg
acgh
acghz
acghzz
acgz
acgzz
ach
achz
achzz
acz
aczz
ad
adc
adcg
adcgh
adcghz
adcghzz
adcgz
adcgzz
adch
adchz
adchzz
adcz
adczz
add
addc
addcg
addcgh
addcghz
addcghzz
addcgz
addcgzz
addch
addchz
addchzz
addcz
addczz
addd
adddc
adddcg
adddcgh
adddcghz
adddcghzz
adddcgz
adddcgzz
adddch
adddchz
adddchzz
adddcz
adddczz
adddg
adddgh
adddghz
adddghzz
adddgz
adddgzz
adddh
adddhz
adddhzz
adddz
adddzz
addg
addgh
addghz
addghzz
addgz
addgzz
addh
addhz
addhzz
addz
addzz
adg
adgh
adghz
adghzz
adgz
adgzz
adh
adhz
adhzz
adz
adzz
ag
agh
aghz
aghzz
agz
agzz
ah
ahz
ahzz
az
azz
c
cg
cgh
cghz
cghzz
cgz
cgzz
ch
chz
chzz
cz
czz
d
dc
dcg
dcgh
dcghz
dcghzz
dcgz
dcgzz
dch
dchz
dchzz
dcz
dczz
dd
ddc
ddcg
ddcgh
ddcghz
ddcghzz
ddcgz
ddcgzz
ddch
ddchz
ddchzz
ddcz
ddczz
ddd
dddc
dddcg
dddcgh
dddcghz
dddcghzz
dddcgz
dddcgzz
dddch
dddchz
dddchzz
dddcz
dddczz
dddg
dddgh
dddghz
dddghzz
dddgz
dddgzz
dddh
dddhz
dddhzz
dddz
dddzz
ddg
ddgh
ddghz
ddghzz
ddgz
ddgzz
ddh
ddhz
ddhzz
ddz
ddzz
dg
dgh
dghz
dghzz
dgz
dgzz
dh
dhz
dhzz
dz
dzz
g
gh
ghz
ghzz
gz
gzz
h
hz
hzz
z
zz