题目

count-and-say

算法

O（n）

• 对于前一个数，找出相同的元素个数，把个数和该元素存入新的string

代码

class Solution {
public:
    string countAndSay(int n) {
        if (n <= 0) return NULL;
        string res = "1";
        for (int i = 1; i < n; ++i) {
            string tmp;
            res.push_back('$'); // Add extra char to deal with boundary
            int count = 0;
            int len = strlen(res.c_str());
            for (int j = 0; j < len; ++j) {
                if (j == 0) ++count;
                else {
                    if (res[j] != res[j - 1]) {
                        tmp.push_back(count + '0');
                        tmp.push_back(res[j - 1]);
                        count = 1;
                    } else ++count;
                }
            }
            res = tmp;
        }
        return res;
    }
};