递归的应用实战一

列一： 斐波拉契数列





strlen递归解法

传统解法：

int strlen(char* s)

{

int n = 0;

while(*s)

{

n++;

s++;

}

return n;

}

递归解法







递归实战应用二







代码：

#include <stdio.h>

#define N 8

typedef struct _tag_Pos

{

int ios;

int jos;

} Pos;

static char board[N+2][N+2];

static Pos pos[] = { {-1, -1}, {-1, 0}, {-1, 1} };

static int count = 0;

void init()

{

int i = 0;

int j = 0;

for(i=0; i<N+2; i++)

{

board[0][i] = '#';

board[N+1][i] = '#';

board[i][0] = '#';

board[i][N+1] = '#';

}

for(i=1; i<=N; i++)

{

for(j=1; j<=N; j++)

{

board[i][j] = ' ';

}

}

}

void display()

{

int i = 0;

int j = 1;

for(i=0; i<N+2; i++)

{

for(j=0; j<N+2; j++)

{

printf("%c", board[i][j]);

}

printf("\n");

}

}

int check(int i, int j)

{

int ret = 1;

int p = 0;

for(p=0; p<3; p++)

{

int ni = i;

int nj = j;

while( ret && (board[ni][nj] != '#') )

{

ni = ni + pos[p].ios;

nj = nj + pos[p].jos;

ret = ret && (board[ni][nj] != '*');

}

}

return ret;

}

void find(int i)

{

int j = 0;

if( i > N )

{

count++;

printf("Solution: %d\n", count);

display();

getchar();

}

else

{

for(j=1; j<=N; j++)

{

if( check(i, j) )

{

board[i][j] = '*';

find(i+1);

board[i][j] = ' ';

}

}

}

}

int main()

{

init();

find(1);

return 0;

}