問題描述:
遞歸是計算機科學中最偉大的思想。按照我的理解,所謂遞歸就是把問題化約爲比自身維度更小的問題,直至邊界點(base condition),
然後利用邊界點的解的結果(相對容易得到)和一定規則回退得到最終所需要的結果。
常用到的遞歸思想的可歸類爲:
1. 各種tree construct 的操作:遍歷(inorder, preorder, postorder),深度優先搜索(dfs),廣度優先搜索(bfs),插入,刪除,返回最值等;
2. 各種搜索問題最終都能化約爲遞歸問題。比如八皇后,排列,組合,揹包等等;
3. 下面我嘗試用遞歸思想實現一些平時最常用的操作(比如加法,乘法等):
代碼如下:
int max( int a, int b )
{
if( a > b )
return a;
return b;
}
int min( int a, int b )
{
if( a < b )
return a;
return b;
}
int Add( int m, int n )
{
if( 0 == n )
return 0;
return Add( m + 1, n - 1);
}
int Mutiple( int m, int n )
{
if( 1 == n )
{
return m;
}
return Mutiple( m + m, n - 1 )
}
int Subtract( int m, int n )
{
if( 0 == n )
{
return 0;
}
return Subtract( m - 1, n - 1 );
}
int findMax( int* items, int idx, int size )
{
if( idx == size - 1)
return items[idx];
return max( findMax( items, idx + 1, size ), items[idx] );
}
int findMin( int* items, int idx, int size )
{
if( idx == size - 1 )
return items[idx];
return min( findMin( items, idx + 1, size ), items[idx] );
}
int Search( int* items, int idx, int size, int val )
{
if( idx >= size )
return -1;
if( items[idx] == val )
return idx;
return Search( items, idx + 1, size, val );
}
int BinarySearch( int* items, int begin, int end, int val )
{
if( begin > end )
return -1;
int mid = begin + ( ( end - begin ) >> 1 );
if( items[mid] > val )
{
return BinarySearch( items, begin, mid - 1, val );
}
else if( items[mi] < val )
{
return BinarySearch( items, mid + 1, end, val );
}
else
{
return mid;
}
}
int gcd( int n, int m )
{
if( 0 == m )
return n;
return gcd( m, n % m );
}
int fibonical( int n )
{
if( 1 == n )
return 1;
return fibonical( n - 1 ) + fibonical( n - 2 );
}
int fictional( int n )
{
if( 1 == n )
return n;
return n * fictional( n - 1 );
}
unsigned int power( int n, int m )
{
if( 1 == m )
return n;
return power( n * m, m - 1 );
}
unsigned int powerQuick( int n, int m )
{
if( 1 == m )
return n;
unsigned int k = powerQuick( n, m / 2 );
if( 0 == m % 2 )
{
return k * k;
}
else
{
return k * k * powerQuick( n, 1 );
}
}
unsigned int Fib( int n, unsigned int a, unsigned int b )
{
if( 1 == n )
return b;
return Fib( n - 1, b, a + b );
}
void EulerLetter( int n, int depth, int* result )
{
if( depth == n )
{
for( int i = 0; i < dpeth; i++ )
{
printf( "%d ", result[i] );
}
printf( "\n" );
return;
}
for( int i = 0; i < n; i++ )
{
if( depth != i )
{
result[depth] = i;
EulerLetter( n, depth + 1, result );
}
}
}