LeetCode刷题笔记--38. Count and Say

最新推荐文章于 2021-03-17 11:00:30 发布

vivian0239

最新推荐文章于 2021-03-17 11:00:30 发布

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本文链接：https://blog.youkuaiyun.com/vivian0239/article/details/88030984

本文探讨了计数与说数序列的生成算法，通过一个C++实现的示例，详细解析了如何根据输入整数n生成序列的第n项。文章深入分析了算法的逻辑，包括字符串操作和循环迭代过程。

38. Count and Say

Easy

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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" or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"

Example 2:

Input: 4
Output: "1211"

不是说好的easy吗，为什么我调试了这么久，还放到VS里了。

class Solution {
public:
   string countAndSay(int n) {
   if(n == 1)return "1";
   string ans = "";
   int calnum = 0;
   string a = "1";
   while (n - 1 > 0)
   {
       ans = "";
       calnum = 0;
       //for (int i = 0; i < a.length(); i++)
       //{
           int j = 0;
           int p = 0;
           char t = a[p];
           char ts = a[j];
           while (j < a.length())
           {
               if (a[j] == a[p])
               {
                   calnum++;
                   j++;
                   if (j == a.length())
                   {
                       ans += t;
                       ans += to_string(calnum);
                   }
               }
               else
               {
                   ans += t;
                   ans += to_string(calnum);
                   calnum = 0;
                   p = j;
                   t = a[p];