[LeetCode] 38. Count and Say 计数和读法

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 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"

Input Constraints:

1 <= n <= 30

对于前一个数，每一段相同元素的子数列数出相同元素的个数，然后这个子数组变为个数+数字，重复到结束。

解法: 迭代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;
    }
};

原文地址：https://www.cnblogs.com/lightwindy/p/9625347.html