【leetcode】38. Count and Say（c/c++，easy难度）

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原题链接 easy

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, 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"

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先解释一下题目，题意有点不清：

n=1时， 数为1

n》2时，开始算下一个数，规则是从左到右，先对一串相同的数字计数再报数

n=2，对1计数是1个1=》11

n=3，对11计数是2个1=》21

n=3，对21计数是1个2 1个1=》1211。

以此类推

思路：

蛮简单的就是遍历用一个count计数。两个string字符串pre、next记录前一个和后一个计数结果

class Solution {
public:
    string countAndSay(int n) {
        if(n == 1) return "1";
       string pre = "1";
       string next = "1";
       char ss[12];
       int i,count = 1;
       int len;

    for(i = 1;i < n;i++){
        len = pre.size();
        next = "";
       
        for(int j = 0;j < len; j++){
            if(pre[j] == pre[j+1] && (j+1) < len) count++;
            else{
                
                sprintf(ss,"%d",count);
                next.append(ss);
                next += pre[j];
                count = 1;
            }
        }
        pre = next;
    }
    return pre;
    }
};