LeetCode 37. Count and Say

n-1為迭代的次數，字串num初始為"1".

每次迭代就開始數字串中數位個數，有一個trick是，當 當前數字與前一位（高位）不同時就停止計數：

所以1211的下一個字串應該是111221(1個1, 1個2, 2個1), 而不是1231(而不是1個2, 3個1).

代碼：

class Solution {public:string countAndSay(int n) {string num("1"), next_num;for (int i = 1; i < n; ++ i){int cnt = '0', last = num[0];next_num.clear();for (size_t j = 0; j < num.size(); ++ j){if (num[j] == last){++ cnt;} else{next_num += cnt;next_num += last;cnt = '1';last = num[j];}}next_num += cnt;next_num += last;num = next_num;}return num;}};