Count and Say

第二篇~~~

这道题比较简单，很容易做出来，下面先说说我的做法~~

题目如下

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.

我的方法是用了两个循环，在第二个循环中用step计算数字重复次数，遇到不重复的数字则跳出同时step置为1，同时更新结果字符串。此过程重复n次后得到最终结果。

public String countAndSay(int n) {
        String temp = "";
        String seq = "1";
        char[] str2char;
        int i = 1;
        int step = 0;
        while(i != n) {
        	temp = "";
        	str2char = seq.toCharArray();
        	for(int p = 0 ; p < str2char.length; p += step) {
        		step = 1; 
        		for(int q = p + 1 ; q < str2char.length ; ++q) {
        			if(str2char[p] == str2char[q])
        				step++;
        			else
        				break;
        		}
        		temp += step + "" + str2char[p];
        	}
        	i++;
        	seq = temp;
        }
        
        return seq;
}

以上是第二篇~~道路仍在继续~~

还有不到5天~~但是我已经开始想你了~我跟你说过的话不只是说说而已~我确实在努力~相信我！~~