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.

Solution:

public class Solution {
    public String countAndSay(int n) {
        char[] seq=new char[100000];
        char[] bak=new char[100000];
        char[] tmp;
        char t;
        int top=1,index,l,r,num;
        seq[0]='1';seq[1]=0;
        while(--n >0){
            index=0;
            for(int i=0;i<top;i++){
                num=1;
                while(i+1<top && seq[i+1]==seq[i]){i++;num++;}
                l=index;
                while(num>0){
                    bak[index++]=(char)(num%10+'0');
                    num/=10;
                }
                r=index-1;
                while(l<r){t=bak[l];bak[l]=bak[r];bak[r]=t;l++;r--;}
                bak[index++]=seq[i];
            }
            top=index;
            tmp=seq;seq=bak;bak=tmp;
        }
        return new String(seq,0,top);
    }
}

C语言源代码（0ms）：

char* countAndSay(int n) {
    char* seq=(char*)malloc(sizeof(char)*100000);
    char* bak=(char*)malloc(sizeof(char)*100000);
    char* tmp;
    char t;
    int top=1,i,index,num,l,r;
    seq[0]='1';seq[1]=0;
    while(--n){
        index=0;
        for(i=0;i<top;i++){
            num=1;
            while(i+1<top && seq[i+1]==seq[i]){
                i++;
                num++;
            }
            l=index;
            while(num>0){
                bak[index++]=num%10+'0';
                num/=10;
            }
            r=index-1;
            while(l<r){
                t=bak[l];bak[l]=bak[r];bak[r]=t;l++;r--;
            }
            bak[index++]=seq[i];
        }
        bak[index]=0;
        top=index;
        tmp=seq;seq=bak;bak=tmp;
    }
    free(bak);
    return seq;
}

C++源代码（0ms）：

class Solution {
public:
    string countAndSay(int n) {
        char* seq=(char*)malloc(sizeof(char)*100000);
        char* bak=(char*)malloc(sizeof(char)*100000);
        char t,*tmp;
        int l,r,index,top=1,i,num;
        seq[0]='1';seq[1]=0;
        while(--n){
            index=0;
            for(i=0;i<top;i++){
                num=1;
                while(i+1<top && seq[i+1]==seq[i]){i++;num++;}
                l=index;
                while(num>0){
                    bak[index++]=num%10+'0';
                    num/=10;
                }
                r=index-1;
                while(l<r){t=bak[l];bak[l]=bak[r];bak[r]=t;l++;r--;}
                bak[index++]=seq[i];
            }
            bak[index]=0;
            top=index;
            tmp=seq;seq=bak;bak=tmp;
        }
        return string(seq);
    }
};

Python源代码（64ms）数据量小时可用，数据量大的话参照上面三种：

class Solution:
    # @param {integer} n
    # @return {string}
    def countAndSay(self, n):
        seq=['1'];top=1;
        while n-1>0:
            n-=1;index=0;bak=[]
            i=0
            while i<top:
                num=1
                while i+1<top and seq[i+1]==seq[i]:i+=1;num+=1
                bak.append(chr(num+ord('0')))
                bak.append(seq[i])
                i+=1
            seq=bak;top=len(bak)
        return ''.join(seq)