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.
这道计数和读法问题还是第一次遇到,看似挺复杂,其实仔细一看,算法很简单,就是对于前一个数,找出相同元素的个数,把个数和该元素存到新的string里。代码如下:
class Solution { public: string countAndSay(int n) { if (n <= 0) return ""; string res = "1"; while (--n) { string cur = ""; for (int i = 0; i < res.size(); ++i) { int cnt = 1; while (i + 1 < res.size() && res[i] == res[i + 1]) { ++cnt; ++i; } cur += to_string(cnt) + res[i]; } res = cur; } return res; } };
博主出于好奇打印出了前12个数字,发现一个很有意思的现象,不管打印到后面多少位,出现的数字只是由1,2和3组成,网上也有人发现了并分析了原因 (http://www.cnblogs.com/TenosDoIt/p/3776356.html),前十二个数字如下:
1 1 1 2 1 1 2 1 1 1 1 1 2 2 1 3 1 2 2 1 1 1 3 1 1 2 2 2 1 1 1 1 3 2 1 3 2 1 1 3 1 1 3 1 2 1 1 1 3 1 2 2 1 1 3 2 1 1 3 1 1 1 2 3 1 1 3 1 1 2 2 1 1 1 1 1 3 1 2 2 1 1 3 3 1 1 2 1 3 2 1 1 3 2 1 2 2 2 1 3 1 1 3 1 1 2 2 2 1 2 3 2 1 1 2 1 1 1 3 1 2 2 1 1 3 1 2 1 1 3 2 1 1
参考资料:
https://discuss.leetcode.com/topic/20195/c-solution-easy-understand