[leetcode 38] Count and Say

Question:

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 nth sequence.

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

分析：

可知道后一个数字为前一个数字的Say；

采用递归的方法，求得前一个数字，则可知道当前数字为多少。

代码如下：

<span style="font-size:14px;">class Solution {public:    string countAndSay(int n) {        string result;        string last;        int count = 1;        if(n == 1)            return "1";        if(n == 2)            return "11";        last = countAndSay(n-1);        for(int i = 1; i < last.size(); ++i){            if(last[i-1] == last[i]){                ++count;                if(i == last.size()-1){                    switch(count){                    case 1:                        result += "1";                        break;                    case 2:                        result += "2";                        break;                    case 3:                        result += "3";                        break;                    case 4:                        result += "4";                        break;                    case 5:                        result += "5";                        break;                    case 6:                        result += "6";                        break;                    case 7:                        result += "7";                        break;                    case 8:                        result += "8";                        break;                    case 9:                        result += "9";                        break;                }                result += last[i-1];              }              continue;            }            else{                switch(count){                    case 1:                        result += "1";                        break;                    case 2:                        result += "2";                        break;                    case 3:                        result += "3";                        break;                    case 4:                        result += "4";                        break;                    case 5:                        result += "5";                        break;                    case 6:                        result += "6";                        break;                    case 7:                        result += "7";                        break;                    case 8:                        result += "8";                        break;                    case 9:                        result += "9";                        break;                }                result += last[i-1];                count = 1;                if(i == last.size()-1){                    result += "1";                    result += last[i];                }            }        }        return result;    }};</span>