LeetCode 38 Count and Say（字符串生成）

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, generate the n^th term of the count-and-say sequence.

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

Example 1:

Input: 1
Output: "1"

Example 2:

Input: 4
Output: "1211"

题目大意：给出一个初始字符串"1"，根据以下规则生成字符串。如果一个字符s[i]与s[i-1]和s[i+1]不同，那么就生成"1"+s[i]的字符串；如果连续n个字符都相同，那么就生成"n"+s[i]的字符串。例如字符串"1"会生成"11"，而"11"会生成"21"，"21"又会生成"1211"，以此类推。

解题思路：有递归和非递归两种解法，核心思路是一样的，只是实现方式不同。每一次生成的字符串都作为下一次的输入，直到生成第n个字符串为止。

代码如下：

//C++实现
class Solution {
public:
    string countAndSay(int n) {
        return countAndSay("1", 1, n);
        // return _countAndSay(n);
    }
private:
    //递归版本
    string countAndSay(string s, int depth, int n) {
        if(depth == n) return s;
        string ans;
        int cnt = 1;
        for(int i = 0;i < s.length();i++){
            if(s[i] == s[i + 1]){
                cnt++;
            }
            else {
                ans += cnt + '0';
                ans += s[i];
                cnt = 1;
            }
        }
        return countAndSay(ans, depth + 1, n);
    }
    //非递归版本
    string _countAndSay(int n) {
        string tmp = "1";
        string ans;
        int cnt = 1;
        for(int i = 2;i <= n;i++){
            for(int j = 0;j < tmp.length();j++){
                if(tmp[j] == tmp[j + 1]){
                    cnt++;
                }
                else {
                    ans += cnt + '0';
                    ans += tmp[j];
                    cnt = 1;
                }
            }
            tmp = ans;
            ans = "";
        }
        return tmp;
    }
};

//C语言实现
#define maxn 10005
char s[maxn], t[maxn];

char* countAndSay(int n) {
    
    
    strcpy(s, "1");
    strcpy(t, "");
    int cnt = 1;
    char tmp[3];
    for(int i = 2;i <= n;i++){
        int len = strlen(s); 
        for(int j = 0;j < len;j++){
            if(s[j] == s[j + 1]){
                cnt++;
            }else {
                tmp[0] = cnt + '0';
                tmp[1] = s[j];
                tmp[2] = '\0';
                strcat(t, tmp);
                cnt = 1;
            }
        }
        strcpy(s, t);
        strcpy(t, "");
    }
    
    return s; 
}