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.
题解:简单的模拟题,但是感觉题目叙述的不是特别清楚。
一开始给定一个“1”,把它说出来就是“one one”,写成串就是11,这样就得到了第二个串11;
把第二个串说出来就是“two one”,写成串就是21,这样就得到了第三个串21;
把第三个串说出来就是“one two one one”,写成串就是1211,这样就得到了第四个串1211;
......
题目求的是第n个串。
设置一个StringBuffer存放当前串“说出来的结果”,然后根据n的大小不断循环,最终得到第n个串即可。
代码如下:
1 public class Solution { 2 public String countAndSay(int n) { 3 String accumu = "1"; 4 5 while(--n > 0){ 6 StringBuffer sb = new StringBuffer(); 7 char[] faster = accumu.toCharArray(); 8 for(int i = 0;i < faster.length;){ 9 int count = 1; 10 char now = faster[i]; 11 i++; 12 while(i<faster.length && faster[i]== faster[i-1] ){ 13 count++; 14 i++; 15 } 16 sb.append(String.valueOf(count)+now); 17 } 18 accumu = sb.toString(); 19 } 20 return accumu; 21 } 22 }
【leetcode刷题笔记】Count and Say
时间: 2024-10-04 07:28:58