countAndSay(n) is the way you would "say" the digit string from countAndSay(n-1), which is then converted into a different digit string. To determine how you "say" a digit string, split it into
leetcode (Count and Say) Title:Count and Say 38 Difficulty:Easy 原题leetcode地址:https://leetcode.com/problems/count-and-say/ 1. 定义变量count,如果相邻的来给你个数相等,那么count++。否则将count和这个数存放,注意最后一个数 时间复杂度:O(n^2),嵌套for循环。 空间复... LeetCode——Count ...
To generate then term, justcount and saythen-1 term. 总之就是第n条结果依赖于第n-1条结果。例如: 结果3=21,结果4=1211,从结果3读出1个2和1个1,所以结果4=1211。 话不多说直接上代码吧: 解题代码: classSolution{public:stringcountAndSay(intn){intcount,getnum; string res="1";if(n==1)retu...
【复试上机】LeetCode-简单-012-Count and Say 12.Count and Say 前言: 这种题还是paper tiger类型,属于文字描述特别多,但是你提取出规律之后就是臭弟弟难度。 文字思想: 递归实现: 结束条件:n==1时,返回”1”。 拆分方法: a)获取上一行的结果,也就是传参为n-1。 b)对上一行进行描述:因为描述的值是连续...
今天介绍的是LeetCode算法题中Easy级别的第12题(顺位题号是38)。count-and-say序列是整数序列,前五个术语如下: 1 11 21 1211 111221 1被读作“一个一”或者11。第二项的值是第一项的读法。 11被读作“两个一”或者21。第三项的值是第二项的读法。
countAndSay(2) = RLE of "1" = "11" countAndSay(3) = RLE of "11" = "21" countAndSay(4) = RLE of "21" = "1211" Example 2: Input:n = 1 Output:"1" Explanation: This is the base case. Constraints: 1 <= n <= 30 ...
append(count > 0 ? count + String.valueOf(cur) : ""); return sb.toString(); } 总结一下 Leetcode上的题目描述我看了好一会儿都没有看懂,后面查了一下,理解了题意,很快就做出来了。 所以理解清楚题目的描述很重要,同时也要学会从题目描述挖掘出信息。 欢迎大家交流,记得点赞~~...
Can you solve this real interview question? Count and Say - The count-and-say sequence is a sequence of digit strings defined by the recursive formula: * countAndSay(1) = "1" * countAndSay(n) is the run-length encoding of countAndSay(n - 1). Run-leng
算法: public String countAndSay(int n) { String r = "1"; for (int i = 1; i < n; i++) { String t = ""; int count = 0; String flag = r.charAt(0) + ""; for (int j = 0; j < r.length(); j++) { if ((r.charAt(j) + "").equals(flag)) { ...
Title:Count and Say 38 Difficulty:Easy 原题leetcode地址:https://leetcode.com/problems/count-and-say/ 1. 定义变量count,如果相邻的来给你个数相等,那么count++。否则将count和这个数存放,注意最后一个数 时间复杂度:O(n^2),嵌套for循环。 空间复... ...