LeetCode——38. Count and Say(C++)

LeetCode——38. Count and Say(C++),第1张

LeetCode——38. 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 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 the minimal number of substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

For example, the saying and conversion for digit string “3322251”:

Given a positive integer n, return the nth term of the count-and-say sequence.

Example 1:

Input: n = 1
Output: “1”
Explanation: This is the base case.
Example 2:

Input: n = 4
Output: “1211”
Explanation:
countAndSay(1) = “1”
countAndSay(2) = say “1” = one 1 = “11”
countAndSay(3) = say “11” = two 1’s = “21”
countAndSay(4) = say “21” = one 2 + one 1 = “12” + “11” = “1211”

Constraints:
1 <= n <= 30

题目大概意思就是从1作为第一个串开始读,读出的串作为下一个待读串,读n-1次,返回第n-1次读出的结果。

class Solution {
public:
    string countAndSay(int n) {
        if(n<=0)return "";
        if(n==1)return "1";
        string pre="1",now="";
        n--;
        while(n--)
        {
            now="";
            for(int i=0;i<pre.size();)
            {
                int cnt=1;
                while(i<pre.size()-1&&pre[i]==pre[i+1])
                {
                    cnt++;
                    i++;
                }
                now+=to_string(cnt)+pre[i];
                i++;
            }
            pre=now;
        }
        return now;
    }
};

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原文地址: https://outofmemory.cn/langs/1499176.html

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