[leetcode]Palindrome Partitioning

 

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"]
  ]

好久木有写C++，手好生，写了很久。。。1016 ms过大集合。。。好像有点慢

思路：

1. 2D dp求出字符串s的回文情况，注意区分aba和aa的情况，解决方法是初始化的时候为1，这样就可以把 f(i, j) = f(i+1, j-1) && s[i] == s[j] 和 s[i] == s[j] 统一起来了

2. 剩下的部分用我的DFS模版写了一下，每次到step1中求出来的isPalin里面去搜，搜到1以后，i = j+1 继续递归下去

 

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;
        
    }
    
    
};