Count and Say [LeetCode 38]】的更多相关文章

1- 问题描述 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…
38. Count and Say Problem's Link ---------------------------------------------------------------------------- Mean: 题目意思太晦涩. 1 读出来 就是“1个1” 所以记为“11” 11 读出来 就是“2个1” 所以记为“21” 21 读出来 就是“1个2 1个1” 所以记为“1221” .... analyse: 略. Time complexity: O(N) view code…
题目链接:https://leetcode.com/problems/count-and-say/?tab=Description   1—>11—>21—>1211—>111221—>312211—>….   按照上面的规律进行求解出第n个字符串是什么.   规律:相连的数字有多少个然后添加上这个数字   参考代码:    package leetcode_50; /*** * * @author pengfei_zheng * 按照规律进行求解字符串 */ publ…
https://leetcode.com/problems/count-and-say/ 题目: 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…
The count-and-say sequence is the sequence of integers with the first five terms as following: 1. 1 2. 11 3. 21 4. 1211 5. 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…
题目描述: 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.…
The count-and-say sequence is the sequence of integers with the first five terms as following: 1. 1 2. 11 3. 21 4. 1211 5. 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…
The count-and-say sequence is the sequence of integers with the first five terms as following: 1. 1 2. 11 3. 21 4. 1211 5. 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…
class Solution { public: vector<string> vs_; Solution(){ "); vs_.push_back(t); ; i< ;++i){ ]; t = ""; ,j ; ; j < t1.size() - ; ++j){ ]){ ++cnt; } else{ ] =""; sprintf(s,"%d%c",cnt,t1[j]); t += string(s); cnt…
这道题主要就是求一个序列,题目得意思就是 1 --> 11 --> 21 --> 1211 -->   111221 --> 312211 --> ..... 1个1     2个1     1个2,1个1   1个1,1个2,2个1  3个1,2个2,1个1 依次类推 题目很简单,但是为了得到较好的结果,还是纠结了一段时间 public class countAndSay { public String countAndSay(int n) { String num…