Palindrome Partitioning——回溯算法的又一经典

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",

Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

判断回文，简单的入栈出栈判断，其他的就是简单的回溯了。

class Solution {
private:
    vector<vector<string>> res;
    vector<string> tempRes;
public:
    bool isValid(string str)
    {
        stack<char> stk;
        for(int i=;i<str.size();i++)
            stk.push(str[i]);
        for(int i=;i<str.size();i++ )
        {
            if(str[i]!=stk.top())
                return false;
            else
                stk.pop();
        }
        return true;
    }
    void getRes(string str)
    {
        if(str.size()==)
        {
            res.push_back(tempRes);
            return ;
        }
        string temp;
        for(int i=;i<str.size();i++)
        {
           temp+=str[i];
           if(isValid(temp))
           {
               tempRes.push_back(temp);
               getRes(str.substr(i+));
               tempRes.pop_back();
           }
        }
    }

    vector<vector<string>> partition(string s) {
        getRes(s);
        return res;
    }
};