[LeetCode]--38. Count and Say

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 nth sequence.

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

这个题目理解起来真是太难了，尤其是英文又不好，只能参看下别人的资料，理解下规则。终于理解，题意是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推，写个countAndSay(n)函数返回字符串。

public String countAndSay(int n) {
        if (n == 1) {
            return "1";
        }
        // 递归调用，然后对字符串处理
        String str = countAndSay(n - 1) + "*";// 为了str末尾的标记，方便循环读数
        char[] c = str.toCharArray();
        int count = 1;
        String s = "";
        for (int i = 0; i < c.length - 1; i++) {
            if (c[i] == c[i + 1]) {
                count++;// 计数增加
            } else {
                s = s + count + c[i];// 上面的*标记这里方便统一处理
                count = 1;// 初始化
            }
        }
        return s;
    }

还有一种非递归的算法。这种我感觉比递归算法写起来就困难一些，理解起来其实差不多，双层循环。

public String countAndSay2(int n) {
            String oldString = "1";
            while (--n > 0) {
                StringBuilder sb = new StringBuilder();
                char [] oldChars = oldString.toCharArray();

                for (int i = 0; i < oldChars.length; i++) {
                    int count = 1;
                    while ((i+1) < oldChars.length && oldChars[i] == oldChars[i+1]) {
                        count++;
                        i++;
                    }
                    sb.append(String.valueOf(count) + String.valueOf(oldChars[i]));
                }
                oldString = sb.toString();
            }

            return oldString;
        }