【LeetCode从零单排】No38.CountAndSay
生活随笔
收集整理的這篇文章主要介紹了
【LeetCode从零单排】No38.CountAndSay
小編覺得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.
題目
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.
代碼
public String countAndSay(int n){String lastNum = "1";for(int i = 1; i <= n - 1 ; i++){StringBuilder sb = new StringBuilder();char[] chars = lastNum.toCharArray();char cur = chars[0];int times = 0;for(int j = 0; j < lastNum.length(); j++){if(chars[j] == cur)times ++;else {sb.append(times).append(cur);cur = chars[j];times = 1;}if(j == lastNum.length() - 1){sb.append(times).append(cur);}}lastNum = sb.toString();}return lastNum;}
代碼下載:https://github.com/jimenbian/GarvinLeetCode
/********************************
* 本文來自博客 ?“李博Garvin“
* 轉(zhuǎn)載請(qǐng)標(biāo)明出處:http://blog.csdn.net/buptgshengod
******************************************/
總結(jié)
以上是生活随笔為你收集整理的【LeetCode从零单排】No38.CountAndSay的全部?jī)?nèi)容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: 【LeetCode从零单排】No36Va
- 下一篇: 【LeetCode从零单排】No58.L