LeetCode solution report -- Count and Say

LeetCode solution report -- Count and Say

Question:

The count-and-say sequence is the sequence of integers
Beginning as follows:
1, 11, 21,121 1, 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.
Click to solve the problem: Count and Say

Analysis: the meaning of the question is to generate a set of strings that meet the requirements of the question through the program,
The question is not very difficult. The basic idea is: Take n as an example, scan the n-1 string from left to right, calculate the number of identical numbers until the scan ends! Initial: count = 1. Each time a number is the same, count ++ converts count to a string or character type + the current array character to find the nth string.

Java code: Accepted

public class Solution {    public String countAndSay(int n) {       String newS = "1";       int count = 1;       int i = 1;       while(i < n){           String s = newS;           newS = "";           for(int j = 0;j < s.length();j ++){                if( (j + 1) < s.length() && s.charAt(j) == s.charAt(j + 1)){                    count ++;                }else{                    newS = newS + count + s.charAt(j);                    count = 1;                }            }            i ++;       }       return newS;    }}

Python code Accepted

class Solution(object):    def countAndSay(self, n):        """        :type n: int        :rtype: str        """        i = 1        count = 1        newS = "1"        while i < n:            s= newS            newS = ""            for j in range(len(s)):                if((j + 1) < len(s) and s[j] == s[j + 1]):                    count = count + 1                else:                    newS = newS + str(count) + s[j]                    count = 1            i = i + 1        return newS