LeetCode 题解(230) : Count and Say

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本文介绍了一种生成计数读数序列的方法，通过给定一个整数n，可以生成第n个计数读数序列。该序列从1开始，后续每一项都是前一项的计数读数表示。

题目：

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 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^th sequence.

Note: The sequence of integers will be represented as a string.

题解：

count occurrence of each type of character and make a new string: count1type1count2type2.....

C++版：

class Solution {
public:
    string countAndSay(int n) {
        if(n == 1)
            return "1";
            
        int i = 1;
        string start = "1";
        while(i < n) {
            string end = "";
            string result = "";
            int j = 0;
            while(j < start.length()) {
                char c = start[j];
                end += c;
                j += 1;
                int count = 1;
                while(j < start.length() && start[j] == c) {
                    count++;
                    j++;
                }
                end = to_string(count) + end;
                result += end;
                end = "";
            }
            start = result;
            i++;
        }
        return start;
    }
};

Java版：

public class Solution {
    public String countAndSay(int n) {
        if(n == 1)
            return "1";
        StringBuffer start = new StringBuffer();
        StringBuffer end = new StringBuffer();
        StringBuffer result = new StringBuffer();
        start.append("1");
        int i = 1;
        while(i < n) {
            end.delete(0, end.length());
            result.delete(0, result.length());
            int j = 0;
            while(j < start.length()) {
                char c = start.charAt(j);
                end.append(c);
                int count = 1;
                j++;
                while(j < start.length() && start.charAt(j) == c) {
                    count++;
                    j++;
                }
                end.insert(0, Integer.toString(count));
                result.append(end.toString());
                end.delete(0, end.length());
            }
            start.delete(0, start.length());
            start.append(result.toString());
            i++;
        }
        return start.toString();
    }
}

Python版：

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        if n == 1:
            return "1"
        i = 1
        start = "1"

        while i < n:
            j = 0
            end = ""
            result = ""
            while j < len(start):
                cur = start[j]
                end += cur
                count = 1
                j += 1
                while j < len(start) and start[j] == cur:
                    count += 1
                    j += 1
                end = str(count) + end
                result += end
                end = ""
            start = result
            i += 1
        return start