[LeetCode]Count and Say

标题叙述性说明

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.

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，因为上次字符是11，有2个1。所以输出21。n=4时。因为上次字符串是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。

解题思路

说实话，题目第一遍看的时候真没看懂，扯的什么犊子啊。后面看了一下网上的描写叙述，理解了题意，事实上非常easy，能够归结为两点：

后一个字符串由前一个字符串生成；
对一个字符串连续同样的数进行描写叙述得到下一个字符串，比方连续三个1（即111）用31来表示；

代码

public static String countAndSay(int n) {
		String result = "1";
		String tempStr;
		char c;
		int count;

		if (n == 1)
			return result;

		for (int i = 2; i <= n; i++) {
			int j = result.length() - 1;
			count = 0;
			tempStr = "";
			c = result.charAt(j);
			//从后往前循环
			while (j >= 0) {
				//记录连续字符的个数
				while (j >= 0 && result.charAt(j) == c) {
					count++;
					j--;
				}

				tempStr = count + "" + c + "" + tempStr;//m个c="mc"
				count = 0;//归0。为记录下一个字符做准备
				
				if (j >= 0) {
					c = result.charAt(j);
				}
			}
			result = tempStr;//得到下一个字符串
		}
        
		return result;
	}