人人網2014筆試演算法題匯總

#include <stdio.h>    int judge(int *a, int len, int k, int *num1, int *num2);    int main(int argc, char **argv)  {      int test_array[] = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};      int result = -1;      int num1, num2;      result = judge(test_array, sizeof(test_array) / sizeof(int), 12, &num1, &num2);      if(result == 0)      {          printf("%d\t%d\n", num1, num2);      }      else if(result == -1)      {          printf("can't find");      }      else      {          printf("error");      }  }    int judge(int *a, int len, int k, int *num1, int *num2)  {      int *low = NULL;      int *high = NULL;      int i = 0;      int result = -1;      if(a == NULL || len < 2)      {          return result;      }      if(a[0] >= k)      {          return result;      }      while(a[i] <= k && i < len)      {          i++;      }      low = a;      high = a + i - 1;      while(low  < high)      {          *num1 = *low;          *num2 = *high;          if((*low + *high) == k)          {              result = 0;              break;          }          else if((*low + *high) > k)          {              high--;          }          else if((*low + *high) < k)          {              low++;          }      }      return result;  }  

解法二：

#include <iostream>    using namespace std;    int hash_table[100];    bool judge(int *a, int len, int x)  {      memset(hash_table, 0, sizeof(hash_table));      for (int i=0; i<len; ++i)      {          hash_table[x - a[i]] = 1;      }        for (int i=0; i<len; ++i)      {          if (hash_table[i] == 1)          {              return true;          }      }        return false;  }    int main()  {      int len = 10;      int a[10] = {1, 3, 5, 7, 9, 4, 2, 8, 10, 6};      int x = 19;        if (judge(a, len, x))      {          cout<<"Yes"<<endl;      }      else      {          cout<<"No"<<endl;      }      system("pause");      return 0;  }  

2.給定有n個數的數組a，其中有超過一半的數為一個定值，在不進行排序，不開設額外數組的情況下，以最高效的演算法找出這個數。

#include <iostream>    using namespace std;    int find(int *a, int n)  {      int t = a[0];      int count = 0;      for (int i=0; i<n; ++i)      {          if (count == 0)          {              t = a[i];              count = 1;              continue;          }          else          {              if (a[i] == t)              {                  count++;              }              else              {                  count--;              }          }      }        return t;  }    int main()  {      int n = 10;      int a[10] = {1, 3, 2, 3, 3, 4, 3, 3, 3, 6};        cout<<find(a, n)<<endl;        system("pause");      return 0;  }