38. Count and Say

Posted 2021-01-12 ruruozhenhao

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 where 1 ≤ n ≤ 30, 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"

 

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 have at least 1 occourence of each digit
    
                // get the number of times same digit occurred (be 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.