LeetCode

Count and Say

2013.12.15 03:00

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:

　　Just do as the problem description says, count and say.

　　Time complexity is...?

　　Let's consider two cases:

　　　　11111->51

　　　　12345->1112131415

　　We'll have to calculate all the sequences s[1], s[1], ..., s[n - 1] before getting s[n]. Thus the time complexity is O(len(s[1]) + len(s[2]) + ... + len(s[n -1])), the average rate of growth of s[i] is between (0, 2), thus the time complexity can be roughly O(2^n).

　　Space complexity is O(2^n) too, as string buffer needs extra space.

　　I'm quite sure that n cannot be very large, otherwise it won't run on an OJ (u_u)

Accepted code:

// 1CE, 1AC
#include <cstdio>
using namespace std;

class Solution {
public:
    string countAndSay(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        string res = "1";
        
        for(int i = 1; i < n; ++i){
            res = nextSequence(res);
        }
        
        return res;
    }
private:
    char buf[100];
    string nextSequence(string cur) {
        string res;
        int i, j;
        int len = cur.length();
        
        res = "";
        i = 0;
        while(i < len){
            j = i + 1;
            while(j < len && cur[i] == cur[j]){
                ++j;
            }
            // 1CE here, you can't simply concatenate string with other data type...
            // use sprintf or sstream to do this
            sprintf(buf, "%d%c", j - i, cur[i]);
            res = res + string(buf);
            i = j;
        }
        
        
        return res;
    }
};