题目：
给定一个字符串 s，将 s 分割成一些子串，使每个子串都是回文串。
返回 s 所有可能的分割方案。

class Solution {
public:
    vector<vector<string>> partition(string s)
    {
        vector< vector<int> > dp = isPalindrome(s, 0, s.size());
        vector<vector<string>> result;
        vector<string> strs = {};
        vector<pair<int, int>> lr = {};
        int end = 0;
        for (int i = 0; i < s.size(); ++i) {
            if (dp[0][i] == 1) {
                lr.push_back({ 0, i });
            }
        }
        while (!lr.empty()) {
            int l = lr.back().first, r = lr.back().second;
            lr.pop_back();
            int len = r - l + 1;
            while (r < end) {
                end -= strs.back().size();
                strs.pop_back();
            }
            strs.push_back(s.substr(l, r - l + 1));
            end += len;
            if (r == s.size() - 1) {
                result.push_back(strs);
                end -= strs.back().size();
                strs.pop_back();
            }
            for (int i = r + 1; i < s.size(); ++i) {
                if (dp[r + 1][i] == 1) {
                    lr.push_back({ r + 1, i });
                }
            }
        }
        return result;
    };

private:
    vector<vector<int>> isPalindrome(string &s, int left, int right)
    {
        int len = s.size();
        vector< vector<int> > dp(len, vector<int>(len, 0));
        for (int i = 0; i < len; i++) {
            dp[i][i] = 1;
            if ((i + 1 < len) && s[i] == s[i + 1]) {
                dp[i][i + 1] = 1;
            }
        }
        for (int i = len - 1; i >= 0; i--) {
            for (int j = i + 1; j < len; j++) {
                if ((s[j] == s[i]) && (i + 1 < len) && (j - 1) >= 0 && (i + 2 <= j)) {
                    dp[i][j] = dp[i + 1][j - 1];
                }
            }
        }
        return dp;
    }
};