LeetCode:38.Count and Say
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Count and Say
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.
题目大意
n=1时输出字符串1;n=2时,数上次字符串中的数值个数,因为上次字符串有1个1,所以输出11;n=3时,由于上次字符是11,有2个1,所以输出21;n=4时,由于上次字符串是21,有1个2和1个1,所以输出1211。依次类推,写个countAndSay(n)函数返回字符串。
解题思路
第一种情况:n<0时返回null。
第二种情况:当n=1时,返回1
第三种情况:当n>1时,假设n-1返回的字符串是s,对s的串进行处理理,对不同的数字进行分组比如112365477899,分成11,2,3,6,5,4,77,8,99。最有就2个1,1个2,1个3,1个6,1个5,一个4,2个7,1个8,2个9,就是211213161614271829,返回此结果。
Java Solution
public class Solution { public String countAndSay(int n) { if (n < 1) { return null; } String result = "1"; for (int i = 2; i <=n ; i++) { result = countAndSay(result); } return result; } public String countAndSay(String str) { StringBuilder builder = new StringBuilder(128); int count = 1; for (int i = 1; i < str.length(); i++) { if (str.charAt(i) == str.charAt(i - 1)) { count++; } else { builder.append(count); builder.append(str.charAt(i - 1)); count = 1; } } builder.append(count); builder.append(str.charAt(str.length() - 1)); return builder.toString(); }}
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