LeetCode38. Count and Say（C++/Python）

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最新推荐文章于 2024-10-17 13:10:58 发布

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本文介绍了一种生成计数与说数序列的方法，通过递归和非递归方式，实现了从第1项到第n项的序列生成。递归方法通过回溯前一项并读出数字和连续出现的次数来生成下一项；非递归方法则通过迭代更新序列，同样实现了序列的生成。

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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 nth 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"

递归解法

class Solution {
public:
    string countAndSay(int n) {
        if(n==1)
            return "1";
        string s=countAndSay(n-1),s2="";
        int len=s.length();
        for(int i=0;i<len;i++){
            int count=1;
            while(i<len-1&&s[i]==s[i+1]){
                count++;
                i++;
            }
            s2+=(count+'0');
            s2=s2+s[i];
        }
        return s2;
    }
};

非递归解法

C++

class Solution {
public:
    string countAndSay(int n) {
       string s="1";
       for(int i=1;i<n;i++){
           string temp="";
           int len=s.length();
           for(int j=0;j<len;j++){
               int count=1;
               while(j<len-1&&s[j]==s[j+1]){
                   j++;
                   count++;
               }
               temp+=(count+'0');
               temp+=s[j];
           }
           s=temp;
       }
        return s;
    }
};

Python

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        ans = "1"
        for i in range(1, n):
            cnt, tempans, tempc = 0, "",  ans[0]
            for c in ans:
                if c == tempc:
                    cnt += 1
                else:
                    tempans += str(cnt) + tempc
                    tempc = c
                    cnt = 1
            tempans += str(cnt) + tempc
            ans = tempans
        return ans