leetcode 38. Count and Say
The count-and-say sequence is the sequence of integers with the first five terms as following:
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"
题目大意:输入一个整数n,返回第n个符合规律的数,n=1, 返回“1”, n=2,返回"11".
思路一:迭代
1 class Solution { 2 public: 3 string countAndSay(int n) { 4 string curr_str = "1"; 5 for (int i = 0; i < n - 1; i++) { 6 string buffer = ""; 7 for (int index = 0; index < curr_str.size(); index++) { 8 int cnt = 1; 9 //统计有多少个与当前所指字符相同的字符 10 while (index < curr_str.size() && curr_str[index + 1] == curr_str[index]) { 11 index++; 12 cnt++; 13 } 14 buffer += to_string(cnt); 15 buffer += curr_str[index]; 16 } 17 curr_str = buffer; 18 } 19 return curr_str; 20 } 21 };
思路二:递归
1 class Solution { 2 public: 3 string countAndSay(int n) { 4 if (n == 1) 5 return "1"; 6 string res, tmp = countAndSay(n - 1); 7 int len = tmp.length(); 8 for (int i = 0; i < len; i++) { 9 int cnt = 1; 10 while (i < len && tmp[i + 1] == tmp[i]) { 11 cnt++; 12 i++; 13 } 14 res += to_string(cnt); 15 res += tmp[i]; 16 } 17 return res; 18 } 19 };
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