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

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"

Input Constraints:

1 <= n <= 30

對於前一個數，每一段相同元素的子數列數出相同元素的個數，然後這個子數組變為個數+數字，重復到結束。

解法: 叠代Iteration

Java:

public String countAndSay(int n) {
	if (n <= 0)
		return null;
 
	String result = "1";
	int i = 1;
 
	while (i < n) {
		StringBuilder sb = new StringBuilder();
		int count = 1;
		for (int j = 1; j < result.length(); j++) {
			if (result.charAt(j) == result.charAt(j - 1)) {
				count++;
			} else {
				sb.append(count);
				sb.append(result.charAt(j - 1));
				count = 1;
			}
		}
 
		sb.append(count);
		sb.append(result.charAt(result.length() - 1));
		result = sb.toString();
		i++;
	}
 
	return result;
}　

Java:

public class Solution {
    public String countAndSay(int n) {
	    	StringBuilder curr=new StringBuilder("1");
	    	StringBuilder prev;
	    	int count;
	    	char say;
	        for (int i=1;i<n;i++){
	        	prev=curr;
	 	        curr=new StringBuilder();       
	 	        count=1;
	 	        say=prev.charAt(0);
	 	        
	 	        for (int j=1,len=prev.length();j<len;j++){
	 	        	if (prev.charAt(j)!=say){
	 	        		curr.append(count).append(say);
	 	        		count=1;
	 	        		say=prev.charAt(j);
	 	        	}
	 	        	else count++;
	 	        }
	 	        curr.append(count).append(say);
	        }	       	        
	        return curr.toString();
        
    }
}

Python:

class Solution:
    # @return a string
    def countAndSay(self, n):
        seq = "1"
        for i in xrange(n - 1):
            seq = self.getNext(seq)
        return seq

    def getNext(self, seq):
        i, next_seq = 0, ""
        while i < len(seq):
            cnt = 1
            while i < len(seq) - 1 and seq[i] == seq[i + 1]:
                cnt += 1
                i += 1
            next_seq += str(cnt) + seq[i]
            i += 1
        return next_seq

Python: wo

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        s = ‘1‘
        i = 1
        while i < n:
            count = 1
            curr = s[0]
            news = ‘‘
            for j in xrange(1, len(s)):
                if curr == s[j]:
                    count += 1
                else:
                    news += str(count) + s[j-1]
                    curr = s[j]
                    count = 1
            news += str(count) + s[-1]        
            s = news    
            i += 1    
                
        return s  　　

C++:

class Solution {
public:
    string countAndSay(int n) {
        if (n <= 0) return "";
        string res = "1";
        while (--n) {
            string cur = "";
            for (int i = 0; i < res.size(); ++i) {
                int cnt = 1;
                while (i + 1 < res.size() && res[i] == res[i + 1]) {
                    ++cnt;
                    ++i;
                }
                cur += to_string(cnt) + res[i];
            }
            res = cur;
        }
        return res;
    }
};

[LeetCode] 38. Count and Say 計數和讀法