The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1 2. 11 3. 21 4. 1211 5. 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 where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
Example 1:
Input: 1 Output: "1"
Example 2:
Input: 4 Output: "1211"
class Solution { public String countAndSay(int n) { StringBuilder sb1 = new StringBuilder("1"), sb2 = sb1; while(--n > 0){ sb2 = new StringBuilder();//初始化外部循环 int c = 1, i = 0; //c表示charAt(i)重复的次数 char[] ch = sb1.toString().toCharArray(); //开始count的标志 while(i < sb1.length()){ if(i+1 < sb1.length() && ch[i] == ch[i+1]) ++c; //重复就++c else{//不重复或到达边界就将当前char存入sb2,c回归1 sb2.append(c).append(ch[i]); c = 1; } ++i; } sb1 = sb2; } return sb1.toString(); } }
来自leetcode,高效简洁
public String countAndSay(int n) { String res = "1"; //从第一行开始,一行一行产生 while (n > 1) { String temp = ""; for (int i = 0; i < res.length(); i++) { int num = getRepeatNum(res.substring(i)); temp = temp + num + "" + res.charAt(i); //跳过重复的字符 i = i + num - 1; } n--; //更新 res = temp; } return res; } //得到字符 string[0] 的重复个数,例如 "111221" 返回 3 private int getRepeatNum(String string) { int count = 1; char same = string.charAt(0); for (int i = 1; i < string.length(); i++) { if (same == string.charAt(i)) { count++; } else { break; } } return count; }