LeetCode-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.

样例输出：

1.     1
2.     11
3.     21
4.     1211
5.     111221
6.     312211
7.     13112221
8.     1113213211
9.     31131211131221
10.   13211311123113112211

自己的解答：

public String countAndSay(int n) {
if (n == 1) {
return "1";
}
String str = countAndSay(n-1);
StringBuilder sb = new StringBuilder();
int i, count;
for (i = 1; i < str.length(); i++) {
count = 0;
while (i == 0 || (i < str.length() && str.charAt(i) == str.charAt(i-1))) {
count++;
i++;
}
sb.append(count);
sb.append(str.charAt(i-1));
}
return sb.toString();
}

见到一个比较好的解答：

public String countAndSay(int n) {
StringBuilder curr=new StringBuilder("1");
StringBuilder prev;
int count;
char say;
for (int i=1;i<n;i++){
prev=curr;
curr=new StringBuilder();
count=1;
say=prev.charAt(0);

for (int j=1,len=prev.length();j<len;j++){
if (prev.charAt(j)!=say){
curr.append(count).append(say);
count=1;
say=prev.charAt(j);
}
else count++;
}
curr.append(count).append(say);
}
return curr.toString();

}

程序的目的就是为了模拟现实，把手动计算计算机化，就基本是一段不错的思路。。。比较上面的代码，先记录say的数字，然后count，不错。。。。

写代码就是为了模拟现实。。。