38. Count and Say*

38. Count and Say*

​​https://leetcode.com/problems/count-and-say/​​

题目描述

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 term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: 4
Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".

C++ 实现 1

给出第 6 个结果就知道题目啥意思了:

1.     1
2.     11
3.     21
4.     1211
5.     111221
6.     312211

发现用双指针就可以搞定, 计算 ​​S[i, ..., j]​​ 中的相同字符多少个. 下面是使用递归的做法.

class Solution {
public:
    string countAndSay(int n) {
        if (n == 1) return "1";
        auto prev = countAndSay(n - 1);
        string res;
        for (int i = 0; i < prev.size(); ++ i) {
            int j = i;
            while (j + 1 < prev.size() && prev[j] == prev[j + 1]) ++ j;
            auto count = j - i + 1;
            res += std::to_string(count) + prev[i];
            i = j;
        }
        return res;
    }
};

C++ 实现 2

下面是使用迭代的做法:

class Solution {
public:
    string countAndSay(int n) {
        string prev = "1";
        while (--n) {
            string res = "";
            for (int i = 0; i < prev.size();) {
                int j = i;
                while (j + 1 < prev.size() && prev[j + 1] == prev[i])
                    ++ j;
                res += to_string(j - i + 1) + prev[i];
                i = j + 1;
            }
            prev = res;
        }
        return prev;
    }
};