最后一遍-L38-Count and Say
描述：
The count-and-say sequence is a sequence of digit strings defined by the recursive formula:
countAndSay(1) = "1"
countAndSay(n) is the way you would "say" the digit string from countAndSay(n-1), which is then converted into a different digit string.
To determine how you "say" a digit string, split it into the minimal number of substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.
For example, the saying and conversion for digit string "3322251":
Given a positive integer n, return the nth term of the count-and-say sequence.
Example 1:
Input: n = 1
Output: "1"
Explanation: This is the base case.
Example 2:
Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"
Constraints:
1 <= n <= 30
主要就是理解题意，countAndSay(n) 的逻辑就是 针对 count(n-1) 进行读一遍 所以是一个迭代的过程
public class L38 {
public static void main(String[] args) {
System.out.println(countAndSay(3));
System.out.println(countAndSay(4));
System.out.println(countAndSay(5));
System.out.println(countAndSay(6));
}
public static String countAndSay(int n) {
if(n == 1) return "1";
if(1 == n) return "1";
if(2 == n) return "11";
String pre = countAndSay(n-1);
StringBuilder result = new StringBuilder();
char cur = pre.charAt(0);
int repeate = 1;
for (int i = 1; i < pre.length(); i++) {
char tmp = pre.charAt(i);
if(cur == tmp){
repeate++;
}else{
result.append(repeate).append(cur);
cur = tmp;
repeate=1;
}
}
result.append(repeate).append(cur);
return result.toString();
}
}