LeetCode OJ:Palindrome Partitioning

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"]
  ]

アルゴリズム思想:深い暴力検索

class Solution{
public:
	int palindrome(string s,int l,int r){
		while(r&&l<s.length()&&s[l]==s[r]){
			l++;r--;
		}
		if(l<r)return 0;
		return 1;
	}
	void dfs(string str,int k,vector<string> t,vector<vector<string>> &store){
		if(k>str.length()-1){
			store.push_back(t);
			return;
		}
		for(int i=k;i<str.length();i++){
			if(palindrome(str,k,i)){
				t.push_back(str.substr(k,i-k+1));
				dfs(str,i+1,t,store);
				t.pop_back();
			}
		}
	}
	vector<vector<string>> partition(string s){
		vector<vector<string>> store;
		vector<string> t;
		dfs(s,0,t,store);
		return store;
	}
};

ダイナミックプランニング

class Solution {
public:
    vector<vector<string>> partition(string s) {
        const int n=s.size();
        bool p[n][n];
        fill_n(&p[0][0],n*n,false);
        for(int i=n-1;i>=0;--i)
            for(int j=i;j<n;++j)
                p[i][j]=s[i]==s[j]&&((j-i<2)||p[i+1][j-1]);
        
        vector<vector<string>> result[n];
        for(int i=n-1;i>=0;--i){
            for(int j=i;j<n;++j){
                if(p[i][j]){
                    const string palindrome=s.substr(i,j-i+1);
                    if(j+1<n){
                        for(auto v:result[j+1]){
                            v.insert(v.begin(),palindrome);
                            result[i].push_back(v);
                        }
                    }else{
                        result[i].push_back(vector<string>{palindrome});
                    }
                }
            }
        }
        return result[0];
    }
};