38 外观数列（中等）

解题思路：从第一个一次遍历统计每个字符出现的次数以及当前字符，然后将其赋值给变量，最终将变量返回。提供C++和python的官方代码

#
# @lc app=leetcode.cn id=38 lang=python3
#
# [38] 外观数列
#

# @lc code=start
class Solution:
    def countAndSay(self, n: int) -> str:
        prev = '1'
        for i in range(n-1):
            curr = ""
            pos = 0
            start = 0

            while pos < len(prev):
                while pos < len(prev) and prev[pos] == prev[start]:
                    pos += 1
                curr += str(pos-start)+prev[start]
                start = pos
            prev = curr
        return prev

# @lc code=end

        

          
        
        
        

          

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class Solution {
public:
    string countAndSay(int n) {
        string prev="1";
        for (int i=2;i<=n;++i){
            string curr="";
            int start=0;
            int pos=0;

            while(pos<prev.size()){
                while (pos<prev.size()&&prev[pos]==prev[start]){
                    pos++;
                }
                curr+=to_string(pos-start)+prev[start];
                start=pos;
            }
            prev=curr;
        }
        return prev;

    }
};

        

          
        
        
        

          

          AI 代码解读