我想生成5 0s字符串的排列,然后是4 0和1的排列,然后是3 0s和2 1s的排列等?我的代码如下:
#include<stdio.h>
int main(){
int i,j,k,l,s[5];
for(i=0;i<5;i++)
s[i]=0;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
printf("---------------------------------------------\n");
for(i=0;i<5;i++){
for(j=0;j<5;j++)
if(i==j)
s[j]=1;
else
s[j]=0;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
}
printf("---------------------------------------------\n");
for(i=0;i<5;i++){
for(k=0;k<5;k++)
s[k]=0;
s[i]=1;
for(j=i+1;j<5;j++){
s[j]=1;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
for(k=j;k<5;k++)
s[k]=0;
}
}
printf("---------------------------------------------\n");
for(i=0;i<5;i++){
for(j=i+1;j<5;j++){
for(k=0;k<5;k++)
s[k]=0;
s[i]=1;
s[j]=1;
for(l=j+1;l<5;l++){
s[l]=1;
for(k=0;k<5;k++)
printf("%d ",s[k]);
printf("\n");
for(k=l;k<5;k++)
s[k]=0;
}
}
}
}
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所以输出是
0 0 0 0 0
---------------------------------------------
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
---------------------------------------------
1 1 0 0 0
1 0 1 0 0
1 0 0 1 0
1 0 0 0 1
0 1 1 0 0
0 1 0 1 0
0 1 0 0 1
0 0 1 1 0
0 0 1 0 1
0 0 0 1 1
---------------------------------------------
1 1 1 0 0
1 1 0 1 0
1 1 0 0 1
1 0 1 1 0
1 0 1 0 1
1 0 0 1 1
0 1 1 1 0
0 1 1 0 1
0 1 0 1 1
0 0 1 1 1
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输出没问题.但是在我的代码中,我对不同的情况使用不同的for循环.是否可以使用更好的方法,以减少代码的长度?
一种方法如下.该解决方案需要O(n)空间,并且每个输出字符串需要O(n)时间.
#include <stdio.h>
#include <stdlib.h>
char *buf;
// Print combinations of m 1's in a field of n 0/1's starting at s.
void print_combinations(char *s, int n, int m)
{
// If there is nothing left to append, we are done. Print the buffer.
if (m == 0 && n == 0) {
*s = '\0';
printf("%s\n", buf);
return;
}
// Cut if there are more 1's than positions left or negative numbers.
if (m > n || m < 0 || n < 0) return;
// Append a 0 and recur to print the rest.
*s = '0';
print_combinations(s + 1, n - 1, m);
// Now do the same with 1.
*s = '1';
print_combinations(s + 1, n - 1, m - 1);
}
int main(void)
{
int n = 5;
buf = malloc(n + 1);
for (int m = 0; m <= n; m++) {
print_combinations(buf, n, m);
printf("-----\n");
}
return 0;
}
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