?

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"]

  ]
  

好久木有寫(xiě)C++，手好生，寫(xiě)了很久。。。1016 ms過(guò)大集合。。。好像有點(diǎn)慢

思路：

1. 2D dp求出字符串s的回文情況，注意區(qū)分aba和aa的情況，解決方法是初始化的時(shí)候?yàn)?，這樣就可以把 f(i, j) = f(i+1, j-1) && s[i] == s[j] 和 s[i] == s[j] 統(tǒng)一起來(lái)了

2. 剩下的部分用我的DFS模版寫(xiě)了一下，每次到step1中求出來(lái)的isPalin里面去搜，搜到1以后，i = j+1 繼續(xù)遞歸下去

?

    class Solution {

    vector<vector<string>> result;

    

    

public:

    vector<vector<bool>> setIsPalin(string s){

        int N = s.size();

        

        vector<vector<bool>> f(N, vector<bool>(N, 1));

        

        

        for(int i = N-1; i >= 0; i--){

            for(int j = i+1; j < N; j++){

                if(j >= N || i < 0) continue;

                f[i][j] = f[i+1][j-1] && s[i] == s[j];

            }

        }

        

        return f;

    }

    

    

    void part(int i, int j, vector<string> tmp, string &s, vector<vector<bool>> &isPalin){

        if(!isPalin[i][j]) return;

        

        tmp.push_back(s.substr(i, j-i+1));

        

        int N  = s.size();

        if(j == N-1){

            result.push_back(tmp);

            return;

        }

        

        i = j+1;

        for(int j = i; j < s.size(); j++){

            part(i, j, tmp, s, isPalin);

        }

        

        

        

    }



    vector<vector<string>> partition(string s) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        result.clear();

        vector<vector<bool>> isPalin = setIsPalin(s);

        vector<string> tmp;

        

        int i = 0;

        for(int j = i; j < s.size(); j++){

            part(i, j, tmp, s, isPalin);

        }

        

        return result;

        

    }

    

    

};
  

?

?

[leetcode]Palindrome Partitioning