LeetCode 38 Count and Say（计数与报数）

翻译

计数报数序列按如下规律开始递增：
1，11，21，1211，111221，……

1 读作“1个1”或11.
11 读作“2个1”或21.
21 读作“1个2，1个1”或1211.

给定一个整数n，生成第n个序列。

备注：数字序列应该用字符串表示。

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.

代码

其实并不难的一道题，不过因为是英文一直没领悟过来，直到看了别人的解。

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<char>(count + '0') + str[i-1];
                count = 1;
            }
            if(i == len - 1)
                result = result + static_cast<char>(count + '0') + str[i];
        }
        return result;
    }
};

然后自己写了这个……

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;
        }
    }
};