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"]
  ]    
分析：本题可以用两种方法解，一种是DFS搜索所有情况，另一个方式是用动态规划。DFS在搜索的时候会有很多判断某个子字符串是否为palindrome的重叠子问题。
DFS方法代码如下：

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string> > result;
        vector<string> path;
        
        dfs(result, path, s, 0);
        
        return result;
    }
    
    void dfs(vector<vector<string> > &result, vector<string> &path, string s, int start){
        if(start == s.length()){
            result.push_back(path);
            return;
        }
        for(int i = start; i < s.length(); i++){
            string tmp = s.substr(start, i-start+1);
            if(isPalindrome(tmp)){
                path.push_back(tmp);
                dfs(result, path, s, i+1);
                path.pop_back();
            }
        }
    }
    
    bool isPalindrome(string s){
        int l = 0, r = s.length() - 1;
        for(; l <= r && s[l] == s[r]; l++, r--);
        return l > r;
    }
};

 动态规划算法如下：

class Solution {
public:
    vector<vector<string>> partition(string s) {
        int n = s.length();
        
        vector<vector<bool> > f(n, vector<bool>(n, false));
        
        for(int i = 0; i < n; i++)
            for(int j = i; j >= 0; j--){
                if(i == j || j + 1 < i && f[j+1][i-1] && s[j] == s[i] || j + 1 == i && s[j] == s[i])
                    f[j][i] = true;
            }
        
        vector<vector<string> > cache[n+1];//cache[0] indicates an empty vector<string>
        cache[0].push_back(vector<string>());
        for(int i = 0; i < n; i++)
            for(int j = -1; j < i; j++){
                if(f[j+1][i]){
                    for(auto v:cache[j+1]){
                        v.push_back(s.substr(j+1, i-j));
                        cache[i+1].push_back(v);
                    }
                }
            }
        
        return cache[n];
    }
};