[英]C++ Condensing nested for loops
我有這些for循環。
// output all possible combinations
for ( int i1 = 0; i1 <= 2; i1++ )
{
for ( int i2 = 0; i2 <= 2; i2++ )
{
for ( int i3 = 0; i3 <= 2; i3++ )
{
for ( int i4 = 0; i4 <= 2; i4++ )
{
for ( int i5 = 0; i5 <= 2; i5++ )
{
for ( int i6 = 0; i6 <= 2; i6++ )
{
for ( int i7 = 0; i7 <= 2; i7++ )
{
//output created words to outFile
outFile
<< phoneLetters[n[0]][i1]<< phoneLetters[n[1]][i2]
<< phoneLetters[n[2]][i3]<< phoneLetters[n[3]][i4]
<< phoneLetters[n[4]][i5]<< phoneLetters[n[5]][i6]
<< phoneLetters[n[6]][i7]
<< " ";
if ( ++count % 9 == 0 ) // form rows
outFile << std::endl;
}
}
}
}
}
}
}
它看起來很糟糕,但我知道從哪里開始凝聚它們太多了。
有人可以給我一兩個指針,這樣我可以使這個代碼更整潔嗎?
你在七個級別上索引0,1和2。 這可能不是非常有效,但是這個怎么樣:
int i1, i2, i3, i4, i5, i6, i7;
int j;
for (int i = 0; i < 2187; i++)
{
// 0 through 2186 represent all of the ternary numbers from
// 0000000 (base 3) to 2222222 (base 3). The following
// pulls out the ternary digits and places them into i1
// through i7.
j = i;
i1 = j / 729;
j = j - (i1 * 729);
i2 = j / 243;
j = j - (i2 * 243);
i3 = j / 81;
j = j - (i3 * 81);
i4 = j / 27;
j = j - (i4 * 27);
i5 = j / 9;
j = j - (i5 * 9);
i6 = j / 3;
j = j - (i6 * 3);
i7 = j;
// print your stuff
}
或者,根據用戶315052在評論中的建議:
int d[7];
for (int i = 0; i < 2187; i++)
{
int num = i;
for (int j = 6; j >= 0; j--)
{
d[j] = num % 3;
num = num / 3;
}
// print your stuff using d[0] ... d[6]]
}
在一般情況下,您可以使用遞歸:
template <typename Stream, typename Iterator>
void generateNumbers(Stream& stream, Iterator begin, Iterator end) {
if (end - begin == 7) {
for (Iterator p = begin; p < end; p++) {
stream << phoneLetters[n[*p]][*p];
}
stream << " ";
} else {
for (*end = 0; *end <= 2; ++*end)
generateNumbers(stream,begin,end+1);
if (end - begin == 6)
stream << std::endl;
}
}
您可以使用緩沖區向量或普通的舊C數組(兩者都具有足夠的大小)來調用它。
例如:
std::vector<int> buf(7,0);
generateNumbers(std::cout,buf.begin(),buf.begin());
// or
int buf2[7];
generateNumbers(std::cout,buf2,buf2);
但如果你的價值是二元的, 那么PBrando的答案會更好。
我看到James McNellis已經評論過這個解決方案,但這里是:
void phone_combo(int n[], int i[], int d, ostream &ofile, int &count) {
if (d == 7) {
//output created words to outFile
ofile
<< phoneLetters[n[0]][i[0]]<< phoneLetters[n[1]][i[1]]
<< phoneLetters[n[2]][i[2]]<< phoneLetters[n[3]][i[3]]
<< phoneLetters[n[4]][i[4]]<< phoneLetters[n[5]][i[5]]
<< phoneLetters[n[6]][i[6]]
<< " ";
if ( ++count % 9 == 0 ) // form rows
ofile << std::endl;
}
return;
}
for (i[d] = 0; i[d] <= 2; i[d]++) {
phone_combo(n, i, d+1, ofile, count);
}
}
int i[7];
phone_combo(n, i, 0, outFile, count);
之前發布了一個響應,將其減少為單個for循環但由於某種原因被刪除。
for( int i(0); i!= 2187; ++i )
{
outFile
<< phoneLetters[n[0]][(i >> 6) & 0x01]<< phoneLetters[n[1]][(i >> 5) & 0x01]
<< phoneLetters[n[2]][(i >> 4) & 0x01]<< phoneLetters[n[3]][(i >> 3) & 0x01]
<< phoneLetters[n[4]][(i >> 2) & 0x01]<< phoneLetters[n[5]][(i >> 1) & 0x01]
<< phoneLetters[n[6]][i & 0x01]
<< ' ';
if ( ++count % 9 == 0 ) // form rows
outFile << '\n';
}
如果您知道計算每個可能的排列所需的確切迭代次數,這只會起作用。
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