LeetCode 38 Count and Say（計數與報數）

翻譯

計數報數序列按如下規律開始遞增：
1，11，21，1211，111221，……

1 讀作“1個1”或11.
11 讀作“2個1”或21.
21 讀作“1個2，1個1”或1211.

給定一個整數n，生成第n個序列。

備注：數字序列應該用字符串表示。

原文

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.

代碼

其實並不難的一道題，不過因為是英文一直沒領悟過來，直到看了別人的解。

class Solution{
public:
    string countAndSay(int n) {
        if(n == 1)
            return "1";
        string result = "1";
        for(int i = 1; i < n; ++i) {
            result = convert(result);
        }
        return result;
    }
    string convert(string str) {
        int len = str.length();
        if(len == 1) return "1" + str;
        string result;
        int count = 1;
        for(int i = 1; i < len; ++i) {
            if(str[i-1] == str[i]) count++;
            else {
                result = result + static_cast(count + '0') + str[i-1];
                count = 1;
            }
            if(i == len - 1) 
                result = result + static_cast(count + '0') + str[i];
        }
        return result;
    }
};

然後自己寫了這個……

class Solution {
public:
    string countAndSay(int n) {
        if (n == 1) {
            return "1";
        } else {
            string output = countAndSay(n - 1), result = "";
            int index = 0;
            while (index < output.size()) {
                char current = output[index];
                int cursor = index, count = 0;
                while (output[cursor] == current && cursor < output.size()) {
                    cursor++; count++;
                }
                char number = count + '0';
                result += number;
                result += current;
                index += count;
            }
            return result;
        }
    }
};