N个数字的全排列问题(用C语言)

[zhangbin55661]

这个问题我已经想了很久了,可还是没想出来,本人菜鸟一个,请高手指点一下,
越详细越好.谢了.

[vvpisces]

排列组合啊！

int i, j;
for (i = 1; i < 10; i++)
for (j = 0; j < 10; j++)
// 这里排列

[zhangbin55661]

这样的效率太低了啊,N=10的话时间复杂度太大了.

[tianma2005123]

折半？

[yeknight]

#include <stdio.h>
void permutation(char a[], int m, int n)
{
int i;
char t;
if (m<n-1) {
permutation(a, m+1, n);
for (i=m+1;i<n;i++) {
t=a[m]; a[m]=a[i]; a[i]=t;
permutation(a, m+1, n);
t=a[m]; a[m]=a[i]; a[i]=t;
}
} else
{
printf("%s ", a);
}
}
int main() {
char a[]="ABCDE";
permutation(a, 0,5);
return 0;
}

[xk306]

#include <stdio.h> 
#define MAX 10 
int used[MAX]; 
int result[MAX]; 
int N; 

void print()
	{ 
		int i; 
		for(i = 0; i < N; i++) 
		printf("%d ", result[i]); 
		printf("\n"); 
	} 
void proc(int step)
	{ 
		int i; 
		if(step == N) 
		print(); 
		else
			{ 
				for(i = 0; i < N; i++)
					{ 
						if(!used[i])
							{ 
								used[i] = 1; 
								result[step] = i + 1; 
								proc(step + 1); 
								used[i] = 0; 
							} 
					} 
			} 
} 

int main()
	{ 
		scanf("%d", &N); 
		proc(0); 
	} 

[xk306]

别人的，借来看看

[scrutin]

百度一下就知道了 可以用回溯法

[九桔猫]

//全排列函数
void proPerm(char *list,int i,int n)
{
    int j,temp;
    if(i==n)
      {
        for(j=0;j<=n;j++)
            printf("%c",list[j]);
            printf("%\n");
      }
    else
      {
         for(j=i;j<=n;j++)
         {
             SWAP(list[i],list[j],temp);
             cout<<i<<j<<endl;
             proPerm(list,i+1,n);
             SWAP(list[i],list[j],temp);
         }
      }
}

其中SWAP是交换两个元素的宏或函数。

[zhangbin55661]

好东西,谢了.

[baihacker]

'我没有C++环境了...用vbs凑合一下
'生成所有排列
Option Explicit

Dim result, counter
const N = 4
const Format = 10

Sub Swap(byRef InArray, first, second)

	Dim t
	t = InArray(first)
	InArray(first) = InArray(second)	
	InArray(second) = t

End Sub

Function GenPer(byRef counter, N, Format)

	Dim c(10), o(10) ,t(10)
	Dim I, j, s, q, result
	Dim oLoopFlag, iLoopFlag

	For I = 0 To N - 1
	    c(I) = 0
	    o(I) = 1
	    t(I) = I
	Next

	oLoopFlag = True
	counter = 0

	While oLoopFlag

		For I = 0 To N - 1
			result = result & t(I)
		Next

		result = result & " "
		counter = counter + 1

		If counter Mod Format = 0 Then
			result = result & chr(13) & chr(10)
		End If

		j = N - 1
		s = 0
		iLoopFlag = True

		While iLoopFlag
			q = c(j) + o(j)
			If q > j or q < 0 Then
				If j = 0 Then
					iLoopFlag = False
					oLoopFlag = False
				Else 	
					If q > j Then
					s = s + 1
					End If
					o(j) = -o(j)
					j = j - 1
				End If
			Else
				Swap t,  j - c(j) + s, j - q + s
				c(j) = q
				iLoopFlag = False
			End If
		Wend

	Wend

	Genper = result

End Function

result = GenPer(counter, N, Format)
result = result & chr(13) & chr(10) & "总数:" & counter

MsgBox(result)