三个抽象建模问题：

• 选择合理数据结构
• 分析模型中内在规律

面试题43：n个骰子点数

面试题44：扑克牌的顺子

 http://www.nowcoder.com/practice/762836f4d43d43ca9deb273b3de8e1f4?tpId=13&tqId=11198&rp=3&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

class Solution {public:    bool IsContinuous( vector
     
       numbers ) {        int len = numbers.size();        sort(numbers.begin(),numbers.end());        if(len <= 0)            return false;        int zero_count = 0;        int gap_count = 0;        for(int i = 0; i < len; i++)        {            if(numbers[i] == 0)                zero_count++;            else if( i + 1 < len)            {                if(numbers[i] == numbers[i+1])                    return false;                else                    gap_count += numbers[i+1] - numbers[i] - 1;            }        }        if(gap_count > zero_count)            return false;        else            return true;    }};
     

 

面试题45：圆圈中最后剩下的数字

用list当作环形链表，当iterator遍历到end，设为begin

注意迭代器不能用四则运算，只能用自增自减，故在第22行用++，然后在26行要还原

1 int LastRemaining_Solution1(unsigned int n, unsigned int m) 2 { 3     if(n < 1 || m < 1) 4         return -1; 5  6     unsigned int i = 0; 7  8     list
     
       numbers; 9     for(i = 0; i < n; ++ i)10         numbers.push_back(i);11 12     list
      
       ::iterator current = numbers.begin();13     while(numbers.size() > 1)14     {15         for(int i = 1; i < m; ++ i)16         {17             current ++;18             if(current == numbers.end())19                 current = numbers.begin();20         }21 22         list
       
        ::iterator next = ++ current;23         if(next == numbers.end())24             next = numbers.begin();25 26         -- current;27         numbers.erase(current);28         current = next;29     }30 31     return *(current);32 }