Count and Say

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 nth sequence.

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

Subscribe to see which companies asked this question

题意：

数数并打印

题目比较不直观，这里的描述比较好一些  http://www.careercup.com/question?id=4425679

"Count and Say problem" Write a code to do following: 
n String to print 
0 1 
1 1 1    因为前面一行有1个1
2 2 1     因为前面一行有2个1
3 1 2 1 1  因为前面一行有1个2和1个1
... 
Base case: n = 0 print "1" 
for n = 1, look at previous string and write number of times a digit is seen and the digit itself. In this case, digit 1 is seen 1 time in a row... so print "1 1" 
for n = 2, digit 1 is seen two times in a row, so print "2 1" 
for n = 3, digit 2 is seen 1 time and then digit 1 is seen 1 so print "1 2 1 1" 
for n = 4 you will print "1 1 1 2 2 1" 

Consider the numbers as integers for simplicity. e.g. if previous string is "10 1" then the next will be "1 10 1 1" and the next one will be "1 1 1 10 2 1"

思路：

遍历，每次对前面的string进行分析，输出

代码：