[leetcode] Count and Say

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1is read off as"one 1"or11.
11is read off as"two 1s"or21.
21is read off as"one 2, thenone 1"or1211.

Given an integer n, generate the n^th sequence.

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

https://oj.leetcode.com/problems/count-and-say/

思路：模拟题，一遍一遍推即可。

public class Solution {
    public String countAndSay(int n) {
        if (n <= 0)
            return null;
        StringBuilder sb = new StringBuilder();
        sb.append("1");
        for (int i = 0; i < n - 1; i++) {
            StringBuilder nsb = new StringBuilder();

            for (int j = 0; j < sb.length();) {
                char cur = sb.charAt(j);
                int cnt = 0;
                while (j<sb.length()&&sb.charAt(j) == cur) {
                    j++;
                    cnt++;
                }
                nsb.append((char)(cnt + ‘0‘));
                nsb.append(cur);

            }

            sb = nsb;
        }
        return sb.toString();

    }

    public static void main(String[] args) {
        System.out.println(new Solution().countAndSay(1));
        System.out.println(new Solution().countAndSay(2));
        System.out.println(new Solution().countAndSay(3));
        System.out.println(new Solution().countAndSay(4));
        System.out.println(new Solution().countAndSay(5));
        System.out.println(new Solution().countAndSay(6));
    }

}

[leetcode] Count and Say,布布扣,bubuko.com

[leetcode] Count and Say

原文：http://www.cnblogs.com/jdflyfly/p/3810747.html