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. Example Given n = 5, return "111221". Note The sequence of integers will be represented as a string.
题解
题目大意是找第 n 个数(字符串表示),规则则是对于连续字符串,表示为重复次数+数本身。
后一个数等于前一个数的计数(字符串表示),有点类似斐波那契数列
JAVA:
public class Solution { /** * @param n the nth * @return the nth sequence */ public String countAndSay(int n) { if (n <= 0) return null; String s = "1"; for (int i = 1; i < n; i++) { int count = 1; StringBuilder sb = new StringBuilder(); int sLen = s.length(); for (int j = 0; j < sLen; j++) { if (j < sLen - 1 && s.charAt(j) == s.charAt(j + 1)) { count++; } else { sb.append(count + "" + s.charAt(j)); // reset count = 1; } } s = sb.toString(); } return s; } }
源码分析
字符串是动态生成的,故使用 StringBuilder 更为合适。注意s 初始化为"1", 第一重 for循环中注意循环的次数为 n-1.
如果 n 比较少的话,可以先生成部分数,弄成一个数组,直接按数组下标输出就行