【LeetCode】38. Count and Say
题目:
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.
提示:
此题没有什么特别的技巧,根据题目要求的推导公式递推下去就行了。
注意:如果仔细思考其中的规律,不难发现不会有同一个数字连续出现4次及以上的情况(原因不妨自己仔细推敲一下)。因此在将字符出现次数的计数器(int型)追加到string当中去的时候,我们可以将它与'0'相加,以此转换为char型,这样子会比用to_string()方法来的速度更快。
代码:
class Solution { public: string countAndSay(int n) { if (n == 1) return "1"; string str = "11"; string res; for (int i = 2; i < n; ++i) { char cur = str[0]; int count = 1; res = ""; for (int j = 1; j < str.size(); ++j) { if (str[j] == cur) { ++count; } else { res += (count+'0'); res += cur; cur = str[j]; count = 1; } } res += (count+'0'); res += cur; str = res; } return str; } };