一些简单数据结构算法的实现4

全排序

#include<iostream>

#include<cstring>

usingnamespace std;

void swap(int &a,int &b)

{

       int temp=a;

       a=b;

       b=temp;

}

void permutationString(char s[],int start){

       int len=strlen(s);

       if(start==len)

       {

              for(inti=0;i<len;i++)

                     cout<<s[i]<<" ";

              cout<<endl;

       }

       for(int i=start;i<len;i++)

       {

              swap(s[start],s[i]);

              permutationString(s,start+1);

              swap(s[start],s[i]);

       }

 

}

void permutation(char s[])

{

       if(s==NULL)

              return;

       permutationString(s,0);

}

 

int main()

{

       char s[]="abc";

       permutation(s);

}

字符串组合的问题

#include<iostream>

#include<vector>

usingnamespace std;

int num=0;

void combination(char s[],int start,int m,int len,vector<char>&result)

{

       if(s==NULL||len<0)

              return;

       if(m==0)

       {

              num++;

              for(inti=0;i<result.size();i++)

              {

                     cout<<result[i];

              }

              cout<<endl;

              return;

       }

 

       if(start==len)

         return;

       result.push_back(s[start]); //加入

       combination(s,start+1,m-1,len,result);

       result.pop_back();

 

       combination(s,start+1,m,len,result);

}

int main()

{

       char s[]="abc";

       int len=strlen(s);

       vector<char> result;

       for(int i=1;i<=len;i++)

       {

              combination(s,0,i,len,result);

       }

       cout<<num<<endl;

}

两个字符串S1,S2，将s1插入S2的某个位置，判断插入后的s2是否是回文。

bool huiwen(char s[]) {

       int len = strlen(s);

       bool flag = true;

       for (int i = 0, j = len - 1; i < j; i++, j--)

              if (s[i] != s[j])

                     flag = false;

       return flag;

}

int main() {

       char s1[100];

       char s2[100];

       char temp[200];

       memset(temp,0,sizeof(temp));

       cin >> s1 >> s2;

       int len1 = strlen(s1);

       int len2 = strlen(s2);

       bool totalflag=false;

       for (int i = 0; i <= len2; i++) {

              int k = i;

              strcpy(temp, s2);

              strcpy(temp + i, s1);

              //cout<< temp << endl;

              for (int j = i + len1; j< (len1 + len2); j++)

                            temp[j] =s2[k++];

              cout << temp << endl;

              bool flag =huiwen(temp);

              if(flag==true)

                     totalflag=true;

       }

   cout<<totalflag<<endl;

       return 0;

}

 

求一个整数的组合，如5=1+1+1+1+1=1+2+2=2+3=。。。。=1+4

 

#include<iostream>

#include<vector>

usingnamespace std;

vector<int> result;

intcountnum= 0;

void getPartition(int a[],int start, int left,int len) {

       if (left == 0) {

              countnum++;

              for (int i = 0; i <result.size(); i++) {

                     cout <<result[i] <<" ";

                     //result.pop_back();  //清空

              }

              cout << endl;

              return;

       }

       for (int i = start; i < len; i++) {

              int select = a[i];

              int y = left;

              int cnt = 0;

              while (y >=select) {

                     result.push_back(select);

                     y -= select;

                     getPartition(a,i + 1, y, len);

                     cnt++;

              }

              while (cnt--) {

                     result.pop_back();

              }

       }

 

}

int main() {

       int n;

       while (cin >> n && n) {

              int a[n];

              int num[n];

              for (int i = 0; i < n-1;i++) {

                     a[i] = i + 1;

              }

              getPartition(a, 0, n, n-1);

              cout << countnum<<endl;

       }

}

 

求数组长度

 

首先一定要记住一点：

 

template <class Type,int size>

Type min(const Type (&Array)[size]){//size用来接收数组的长度--元素个数

   Type min_value = Array[0];

   for(int i=1;i<size;++i){

       if(Array[i]<min_value) min_value=Array[i];

   }

   return min_value;

}

 

有了以上的函数模板,然后就可以使用min这个函数了:

 

int main()

{

   float fa[5]={1.1,2.2,3.3,4.4,5.5};

   int ia[]={1,2,3,4,5,6,7,8,9,0};

   float fmin=min(fa);//(1)

   int imin=min(ia);//(2)

   return0;

 

}

 

在(1)处,函数min()就根据模板变成了下面的样子:

float min(constfloat (&Array)[5]){

   float min_value=Array[0];

   for(int i=1;i<5;++i)

   if(Array[i]<min_value) min_value=Array[i];

   return min_value;

}

 

在(2)处,函数min()就根据模板变成了下面的样子:

int min(constint (&Array)[10]){

   int min_value=Array[0];

   for(int i=1;i<10;++i)

   if(Array[i]<min_value) min_value=Array[i];

   return min_value;

}

用位运算实现加法：

#include<iostream>

#include<cstring>

#include<stack>

usingnamespace std;

 

int  add(int  num1,int num2)

{

       int sum=num1^num2;

       int carry=(num1&num2)<<1;

        num1=sum;

        num2=carry;

       while(carry!=0)

       {

              sum=num1^num2;

              carry=(num1&num2)<<1;

              num1=sum;

              num2=carry;

       }

       return num1;

}

int main() {

       cout<<add(3,8)<<endl;

       cout<<add(-3,8)<<endl;

       return 0;

}

 

计算 序列的全排序的代码，用递归

#include<iostream>

usingnamespace std;

int num=0;//the number of list

void printArr(int str[],int len)

{

       for(int i=0;i<len;i++)

              cout<<str[i]<<" ";

       cout<<endl;

}

void swap(int &a,int &b)

{

       int temp=a;

       a=b;

       b=temp;

}

void permutation(int str[],int start,int len)

{

       if(str==NULL||len<=0)

              return ;

       if(start==len)

       {

              printArr(str,len);

              num++;

       }else

       {

              for(intj=start;j<len;j++)

              {

            swap(str[start],str[j]);

           permutation(str,start+1,len);

            swap(str[start],str[j]);

              }

       }

}

int main()

{

    int str[]={1,2,3,4};

    permutation(str,0,4);

    cout<<num<<endl;

}

 

 

 

void Permutation(char* pStr, char* pBegin);

/////////////////////////////////////////////////////////////////////////
// Get the permutation of a string, 
// for example, input string abc, its permutation is 
// abc acb bac bca cba cab
/////////////////////////////////////////////////////////////////////////
void Permutation(char* pStr)
{
      Permutation(pStr,pStr);
}

/////////////////////////////////////////////////////////////////////////
// Print the permutation of a string, 
// Input: pStr   -input string
//        pBegin- points to the begin char of string 
//                 which we want to permutate in thisrecursion
/////////////////////////////////////////////////////////////////////////
void Permutation(char* pStr, char* pBegin)
{
      if(!pStr || !pBegin)
            return;

      // if pBegin pointsto the end of string,
      // this round of permutation isfinished, 
      // print the permuted string
      if(*pBegin == '\0')
      {
            printf("%s\n", pStr);
      }
      // otherwise,permute string
      else
      {
            for(char* pCh = pBegin; *pCh != '\0'; ++ pCh)
            {
                  // swap pCh and pBegin
                  char temp = *pCh;
                  *pCh = *pBegin;
                  *pBegin = temp;

                  Permutation(pStr, pBegin + 1);

                  // restore pCh and pBegin
                  temp = *pCh;
                  *pCh = *pBegin;
                  *pBegin = temp;
            }
      }
}