Leetcode Palindrome Partitioning

題目：

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"]
  ]

解題：

先用dp記錄迴文串，dp[i][j]表示s串從i到j爲迴文，狀態轉移：如果s[i] == s[j]，並且i == j || j == i + 1 || dp[i + 1][j - 1] == 1則標記dp[i][j]爲1，表示是迴文

然後暴力dfs。

代碼如下：

class Solution {
public:
    vector<vector<string> > partition(string s) {
        makeDP(s);
        ans.clear(), single.clear();
        dfs(0, s);
        return ans;
    }
private:
    static const int SIZE = 1000;
    bool dp[SIZE][SIZE];
    vector<vector<string> > ans;
    vector<string> single;

    void makeDP(string s) {
        memset(dp, 0, sizeof(dp));
        for(int i = s.size() - 1; i >= 0; i --)
            for(int j = i; j < s.size(); j ++) 
                if(s[i] == s[j] && (i == j || j == i + 1 || dp[i + 1][j - 1]))
                    dp[i][j] = 1;
    }

    void dfs(int x, string s) {
        if(x == s.size()) {
            ans.push_back(single);
            return ;
        }
        for(int i = x; i < s.size(); i ++) {
            if(dp[x][i]) {
                string tmp;
                tmp.assign(s, x, i - x + 1);
                single.push_back(tmp);
                dfs(i + 1, s);
                single.pop_back();
            }
        }
    }
};