LeetCode.38 Count and Say
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题目:
The count-and-say sequence is the sequence of integers with the first five terms as following
1. 12. 113. 214. 12115. 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 term of the count-and-say sequence.
Note: Each term of the sequence of integers will be represented as a string.
提示:
报数
该题的意思:给定一个数n,返回第n次的字符串。条件为:碰到1个1读作11,碰到11读作2个1,碰到12读作1个1和1个2。
规律为:判断是否连续,统计记数;表示:(次数,数字)。
第n此报数以n-1次的字符串为开始,使用递归实现。
1---> 1
2--->11
3--->21
4--->12115--->111221
6--->312211
7--->13112221
分析:
class Solution { public String countAndSay(int n) { // 使用递归实现if (n == 1)return "1";String str = countAndSay(n - 1)+"/";//为了方便判断最后个数据int count = 1;StringBuilder sb = new StringBuilder();for (int i = 0; i < str.length()-1; i++) {if(str.charAt(i)==str.charAt(i+1)){count++;}else{sb.append(String.valueOf(count)+str.charAt(i));count=1;//初始化}}return sb.toString();}}
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