[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;
}