[LeetCode 471] Encode String with Shortest Length

Trick:  int r = (sub + sub).find(sub, 1); 寻找重复pattern。sub.size() % r == 0一定成立。否则设r是查找结果，总能找到更小的r满足条件。

class Solution {
public:
    string encode(string s) {
        int n = s.size();
        vector<vector<string>> dp(n + 1, vector<string>(n + 1));
        for (int i = 0; i < n; i++) {
            dp[i][i + 1].push_back(s[i]);
        }
        for (int k = 2; k <= n; k++) {
            for (int i = 0, j = k; j <= n; i++, j++) {
                dp[i][j] = dp[i][j - 1] + dp[j - 1][j];
                for (int m = i + 1; m < j; m++) {
                    string tmp = dp[i][m] + dp[m][j];
                    if (tmp.size() < dp[i][j].size()) {
                        dp[i][j] = tmp;
                    }
                }
                int len = j - i;
                string sub = s.substr(i, len);
                int r = (sub + sub).find(sub, 1);
                if (r < len) {
                    string tmp = to_string(len / r) + "[" + dp[i][i + r] + "]";
                    if (tmp.size() < dp[i][j].size()) {
                        dp[i][j] = tmp;
                    }
                }
            }
        }
        return dp[0][n];
    }
};