[leetcode] 38. Count and Say-优快云博客

本文介绍了一种基于迭代方式生成计数并说序列的方法，通过C++实现了一个简单的算法来生成给定整数n对应的序列。该算法利用字符串处理技巧，通过计数和追加字符的方式生成下一个序列。

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.

这道题是迭代生成数字字符串，题目难度为Easy。

题目比较简单，主要在于字符串的处理上，这里用to_string函数将计数转化为字符串，还可以用sprintf或stringstream，其他应该没什么难度了。具体代码：

class Solution {
public:
    string countAndSay(int n) {
        if(n <= 0) return "";
        if(n == 1) return "1";
        
        string curStr = "1", newStr = "";
        for(int i=0; i<n-1; ++i) {
            int cnt = 1;
            for(int j=1; j<curStr.size(); ++j) {
                if(curStr[j] == curStr[j-1]) cnt++;
                else {
                    newStr += to_string(cnt);
                    newStr += curStr[j-1];
                    cnt = 1;
                }
            }
            newStr += to_string(cnt);
            newStr += curStr[curStr.size()-1];
            curStr = newStr;
            newStr = "";
        }
        
        return curStr;
    }
};