2013年01月01日

//求一個字符串中連續出現次數最多的子串。

#include <cstdlib>
#include <vector>
#include <iostream>
#include <string>
using namespace std;

pair<int, string> fun(conststring& str)
{
        vector<string> substrs;
         int maxcount = 1;
         int count = 1;
         string substr;
         int len = str.length();
         for(int i = 0; i< len; i++)
               substrs.push_back(str.substr(i, len-i));
         for(int i = 0; i< len; i++)
         {
                for(int j= i+1; j < len; j++)
                {
                      count = 1;
                      if(substrs[i].substr(0, j-i) ==substrs[j].substr(0, j-i))
                      {
                                          ++count;
                                          for(int k = j+(j-i); k < len; k+= j-i)
                                          {
                                                 if(substrs[i].substr(0, j-i)== substrs[k].substr(0, j-i))
                                                                     count++;
                                                 else
                                                    break;
                                          }
                                          if(count > maxcount)
                                          {
                                                  maxcount =count;
                                                  substr =substrs[i].substr(0, j-i);
                                          }
                      }
                }
         }
         return make_pair(maxcount,substr);
}
int main(int argc, char *argv[])
{
    string str;
    pair<int,string> rs;
    while(cin>> str)
    {
            rs = fun(str);
            cout << rs.second<< ":"<< rs.first<< "\n";
    }
    system("PAUSE");
    returnEXIT_SUCCESS;
}

//編程之美2.12快速找出一個數組中的兩個數字，讓這兩個數字之和等於一個給定的值。首先對數組進行排序，然後遍歷一次。時間複雜度O(n)。

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

bool cmp(int a, int b)
{
     returna < b;
}
void find(int a[], int sum)
{
     int i,j;
     for(j= 6, i = 0; i < j;)
    {
                  if(a[i] + a[j]< sum)
                         i++;
                  else if(a[i] + a[j]> sum)
                      j--;
                  else 
                  {
                      cout << i<< " "<< j <<endl;
                      i++;
                      j--;
                  }
    }
    return;
}

int main(int argc, char *argv[])
{
    int a[] = {5, 6, 1, 4,7, 9, 8};
    int sum = 10;
    sort(a, a+7, cmp);
    for(int i = 0; i< 7; i++)
       cout << a[i]<< " ";
    cout<< endl;
    find(a, sum);
    system("PAUSE");
    returnEXIT_SUCCESS;
}

//編程之美3.3 計算字符串的相似度。

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int min(int t1, int t2, int t3)
{
    if(t1 >t2)
         t1 = t2;
    if(t1 >t3)
         t1 = t3;
    return t1;
}

int calculate(string strA, int Abegin, int Aend, string strB,int Bbegin, int Bend)
{
    if(Abegin> Aend)
    {
            if(Bbegin > Bend)
                     return0;
            else
                returnBend-Bbegin+1;
    }
    if(Bbegin> Bend)
    {
            if(Abegin > Aend)
                     return0;
            else
                returnAend-Abegin+1;
    }
    if(strA[Abegin] ==strB[Bbegin])
                 return calculate(strA, Abegin+1, Aend, strB,Bbegin+1, Bend);
    else
    {
       int t1 = calculate(strA, Abegin, Aend, strB,Bbegin+1, Bend);
       int t2 = calculate(strA, Abegin+1, Aend, strB,Bbegin, Bend);
       int t3 = calculate(strA, Abegin+1, Aend, strB,Bbegin+1, Bend);
       return min(t1, t2, t3)+1;
    }
}

int main(int argc, char *argv[])
{
    string strA ="xabcdae";
    string strB ="xfdfa";
    int m = calculate(strA,0, 7, strB, 0, 5);
    cout<< m <<endl;
    system("PAUSE");
    returnEXIT_SUCCESS;
}

使用以上算法存在的問題：在遞歸的過程中，有數據被重複計算。使用DP，類似於求兩個字符串的最長公共子序列。

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int min(int t1, int t2, int t3)
{
    if(t1 >t2)
         t1 = t2;
    if(t1 >t3)
         t1 = t3;
    return t1;
}

int calculate(string strA, string strB)
{
    int lenA =strA.length()+1;
    int lenB =strB.length()+1;
    int c[100][100];
    for(int i = 1; i< lenA; i++)
           c[i][0] =i;
    for(int j = 1; j< lenB; j++)
           c[0][j] =j;
    c[0][0] = 0;
    for(int i = 1; i< lenA; i++)
           for(int j= 1; j < lenB; j++)
           {
                 if(strB[j-1] == strA[i-1])
                            c[i][j] = c[i-1][j-1];
                 else
                     c[i][j] =min(c[i][j-1], c[i-1][j], c[i-1][j-1])+1;
           }
    for(int i = 0; i< lenA; i++)
    {
           for(int j= 0; j < lenB; j++)
                 cout << c[i][j]<< " ";
           cout<< endl;  
    }
    returnc[lenA-1][lenB-1];
}

int main(int argc, char *argv[])
{
    string strB ="xabcdae";
    string strA ="xfdfa";
    int m = calculate(strA,strB);
    cout<< m <<endl;
    system("PAUSE");
    returnEXIT_SUCCESS;
}