131. Palindrome Partitioning(回文子串划分 回溯)

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"]
]

 

 

class Solution {
public:
    vector<vector<string>> res;
    void backtrack(vector<string>& path,string& s, int start) {
        if (start == s.size()) {
            res.push_back(path);
            return;
        }
        for (int i = start; i < s.size(); ++i) {
            if (!isPalindrome(s,start,i)) {
                continue;
            }
            string ss = s.substr(start,i-start+1);
            path.push_back(ss);
            backtrack(path,s,i+1);
            path.pop_back();
        }
    }
    bool isPalindrome(const string s,int start, int end) {
        for(int i = start,j = end; j>i;++i,--j) {
            if(s[i]!=s[j]) {
                return false;
            }
        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        vector<string> path;
        backtrack(path,s,0);
        return res;
    }
};

 

 

 

 

 1 class Solution {
 2     private List<List<String>> res = new ArrayList<>();
 3     public List<List<String>> partition(String s) {
 4        
 5     help(new ArrayList<>(), s, 0);
 6        return res;
 7     }
 8 
 9     private void help( List<String> temp, String s, int index){
10        if(index == s.length())
11           res.add(new ArrayList<>(temp));
12        else{
13           for(int i = index; i < s.length(); i++){
14              if(isPalindrome(s, index, i)){
15                 temp.add(s.substring(index, i + 1));
16                 help(temp, s, i + 1);
17                 temp.remove(temp.size() - 1);
18              }
19           }
20        }
21     }
22 
23     public boolean isPalindrome(String s, int low, int high){
24        while(low < high)
25           if(s.charAt(low++) != s.charAt(high--)) return false;
26        return true;
27     } 
28 }