LeetCode:Count and Say

Count and Say

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Question

Total Accepted: 62217 Total
Submissions: 233796 Difficulty: Easy

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

.

Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.

题意：

第1个数：1

第2个数：从前一个数来，上一个数是1个1，故：11

第3个数：上一个数是2个1，故：21

第4个数：上一个数是1个2 + 1个1，故：1211

...

求第n个数。

code:

class Solution {
public:
    string countAndSay(int n) {
        
        string s[2];
        s[0]="1";
        s[1]="";
        int cur=0;
        for(int i=1;i<n;i++) {
            
            int cnt;
            for(int j=0;j<s[cur].size();j+=cnt) {
                cnt=0;
                char t=s[cur][j];
                for(int k=j;k<s[cur].size() && t==s[cur][k];k++) cnt++;
                s[cur^1]+='0'+cnt;
                s[cur^1]+=t;
            }
            s[cur]="";
            cur^=1;
        }
        return s[cur];
    }
};