. Count and Say

The Count-and-say sequence is the sequence of integers with the first five terms as following:

1.     .     113.     214.     12115.     111221

1is read off as "one 1" or 11 .
11is read off as "two 1s" or 21 .
21is read off "one 2 as, then one 1" or 1211 .

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

Note:each term of the sequence of integers would be represented as a string.

Example 1:

Input:1output: "1"

Example 2:

Input:4output: "1211"

AC Code:

Class Solution {public:    string Countandsay (int n) {        vector<string> v;        Init (v, n);        return v[n-1];    }    void Init (vector<string>& v, int n) {        v.push_back ("1");        for (int i = 1; I <= n; ++i) {            string res = v[i-1];            char C = res[0];            String ans = "";            int num = 1;            for (int i = 1; i < res.length (); ++i) {                 if (res[i] = = c) {                    num++;                    Continue;                } else {                    ans = ans + to_string (num) + C;                    c = res[i];                    num = 1;                }             }            Ans = ans + to_string (num) + C;            V.push_back (ANS);}}    ;

Runtime:8 MS, faster than 8.70% of C + + online submissions for Count and Say.

Class Solution {public:std::string Countandsay (int n) {if (0 = = N) return "";        if (1 = = N) return "1";    std::string res= "1";        std::string s;                        for (int i = 1; i < n; i++) {//run from starting to generate second string int len = Res.size (); Cheack all digits in the string for (int j = 0; J < Len; J + +) {int count=1;//We Ha  ve at least 1 occourence for each digit//get the number of times same digit occurred (being carefull with    The end of the string) while ((j + 1 < len) && (res[j] = = res[j + 1])) {count++;        j + +; We need to keep increasing the index inside of the string}//Add to New string "count" +    "Digit itself" s + = std::to_string (count) + res[j];        }//Save temporary result res = s;        Clean our string-helper s.clear ();    } return res; }};

Runtime:4 MS, faster than 41.19% of C + + online submissions for Count and Say.

char* Countandsay (int n) {    if (n = = 1) return "1", char *cur = malloc (2), *tmp;cur[0] = ' 1 '; cur[1] = 0;int len, idx, J  , count;for (int i = 2; I <= n; ++i) {len = strlen (cur); tmp = malloc (len * 3); memset (tmp, 0, Len * 3); count = 1;for (idx = 1, j = 0; IDX < Len; ++IDX) {if (cur[idx] = = Cur[idx-1])        {            ++count;        } else        {            tmp[j++] = ' 0 ' + count;            Tmp[j++] = cur[idx-1];            Count = 1;}        } tmp[j++] = ' 0 ' + count;    Tmp[j++] = cur[len-1];free (cur); cur = tmp;} return cur;}

Runtime: 0 ms, faster than 100.00% of C online submissions for Count and Say.

. Count and Say