131. 分割回文串

链接：131. 分割回文串

题解：https://leetcode-cn.com/problems/palindrome-partitioning/solution/hui-su-fa-si-lu-yu-mo-ban-by-fuxuemingzh-azhz/

https://leetcode-cn.com/problems/palindrome-partitioning/solution/hui-su-you-hua-jia-liao-dong-tai-gui-hua-by-liweiw/

https://leetcode-cn.com/problems/palindrome-partitioning/solution/131-fen-ge-hui-wen-chuan-hui-su-sou-suo-yp2jq/

class Solution {
public:
    // 判断从字符串s，[begin, i]的字符串是否为回文串
    bool is_same(const std::string& s, int begin, int i) {
        while(begin <= i) {
            if(s[begin] != s[i]) {
                return false;
            }
            ++begin;
            --i;
        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        if(s.size() == 0) {
            return res;
        }
        int begin = 0; // 截取字符串的开头
        vector<string> path; // 临时路径
        dfs(s, begin, path, res);
        return res;
    }
    void dfs(const std::string& s, int begin, vector<string>& path, vector<vector<string>>& res) {
        if(begin >= s.size()) {
            res.push_back(path);
            return;
        }
        for(int i = begin; i < s.size(); ++i) {
            // 判断当前层字符串[begin,i]是否为回文串
            if(!is_same(s, begin, i)) {
                continue;
            }
            // s[begin,i]为回文串
            path.push_back(s.substr(begin, i-begin+1));
            // 递归下一层
            dfs(s, i+1, path, res);
            // 回溯
            path.pop_back();
        }
    }
};

标签：begin,int,res,分割,hui,131,path,回文
来源： https://blog.csdn.net/INGNIGHT/article/details/114501966