[leetcode] Count and Say

Count and Say

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the n^th sequence.

Note: The sequence of integers will be represented as a string.

 

题意是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所 以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推，写个countAndSay(n)函数返回字符串。

 

分析：将数字转换为字符串，依次统计相同字符s的个数count，然后采用 new_str = new_str + count + s的方式，组成新的字符串。

         以上操作重复n-1次即可。

 

class Solution
{
private:
  string IntToString(int n)
  {
    if(n==0)
      return "0";

    string str = "";
    while(n>0)
    {
      str.insert(str.begin(), n % 10 + '0');
      n /= 10;
    }
    return str;
  }

public:
  string countAndSay(int n)
  {
    if(n == 1)
      return "1";

    if(n == 2)
      return "11";

    int i=0 ,j=0;

    string temp = "11", str = "";
    for(i=3; i<=n; i++)
    {
      int count = 1;
      str = "";
      for(j=1; j<temp.size(); j++)
      {
        if(temp[j] == temp[j-1])
          count++;
        else
        {
          str = str + IntToString(count) + temp[j-1];
          count = 1;
        }
        if(j == temp.size()-1)
          str = str + IntToString(count) + temp[j];
      }
      temp = str;
    }
   
    return str;
  }
};