1. 1 2. 11 3. 21 4. 1211 5. 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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
Example 1:
Input: 1 Output: "1"
Example 2:
Input: 4 Output: "1211"
思路:
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。
1 class Solution { 2 public: 3 string countAndSay(int n) { 4 string res = "1"; 5 6 for (int i = 2; i <= n; i++) 7 res = countsay(res); 8 return res; 9 } 10 11 string countsay(string vec){ 12 13 string res; 14 char key = vec[0]; 15 int cnt = 1; 16 if (vec == "1") 17 return "11"; 18 19 for (int i = 1; i<vec.size(); i++) 20 { 21 if (key == vec.at(i)){ 22 cnt++; 23 24 else 25 { 26 res = res + to_string(cnt) + key; 27 key = vec[i]; 28 cnt = 1; 29 } 30 if (i == vec.size()-1) 31 res = res + to_string(cnt) + key; 32 33 } 34 35 return res; 36 37 } 38 };