131. Palindrome Partitioning

题目：

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"]
  ] 

链接：  http://leetcode.com/problems/palindrome-partitioning/

题解：

一看到return all xxxx，就猜到可能要用回溯。这道题就是比较典型的递归+回溯。递归前要判断当前的子字符串是否palindrome，答案是false的话要continue。

Time Complexity - O(n*2ⁿ)， Space Complexity - O(n*2ⁿ)

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        if(s == null || s.length() == 0)
            return res;
        ArrayList<String> list = new ArrayList<>();
        partition(res, list, s, 0);
        return res;
    }
    
    private void partition(List<List<String>> res, ArrayList<String> list, String s, int pos) {
        if(pos == s.length()) {
            res.add(new ArrayList<String>(list));
            return;
        }
        
        for(int i = pos + 1; i <= s.length(); i++) {
            String partition = s.substring(pos, i);
            if(!isPalindrome(partition))
                continue;
            list.add(partition);
            partition(res, list, s, i);
            list.remove(list.size() - 1);
        }
    }
    
    
    private boolean isPalindrome(String s) {
        int lo = 0, hi = s.length() - 1;
        
        while(lo < hi) {
            if(s.charAt(lo) != s.charAt(hi))
                return false;
            lo++;
            hi--;
        }
        
        return true;
    }
}

需要好好看看主方法来确定定量分析递归算法的时间复杂度。

二刷:

仔细想一想代码可以简化不少。主要分为三部分。1是题目给定的方法，2是辅助方法，用来递归和回溯，3是判断string是否是palindrome。注意考虑清楚需要多少变量，以及时间空间复杂度。

Time Complexity: O(n!)

Space Complexity: O(n ^ 2)

Java:

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        List<String> list = new ArrayList<>();
        partition(res, list, s);
        return res;
    }
    
    private void partition(List<List<String>> res, List<String> list, String s) {
        if (s == null || s.length() == 0) {
            res.add(new ArrayList<String>(list));
            return;
        }
        for (int i = 0; i < s.length(); i++) {
            String subStr = s.substring(0, i + 1);
            if (isPalindrome(subStr)) {
                list.add(subStr);
                partition(res, list, s.substring(i + 1));
                list.remove(list.size() - 1);
            }
        }
    }
    
    private boolean isPalindrome(String s) {
        if (s == null || s.length() < 2) {
            return true;
        }
        int lo = 0, hi = s.length() - 1;
        while (lo <= hi) {
            if (s.charAt(lo) != s.charAt(hi)) {
                return false;
            }
            lo++;
            hi--;
        }
        return true;
    }
}

三刷:

依然是使用二刷的方法。

Java:

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        if (s == null || s.length() == 0) return res;
        partition(res, new ArrayList<>(), s);
        return res;
    }
    
    private void partition(List<List<String>> res, List<String> list, String s) {
        if (s.length() == 0) {
            res.add(new ArrayList<String>(list));
            return;
        }
        for (int i = 0; i <= s.length(); i++) {
            String front = s.substring(0, i);
            if (isPalindrome(front)) {
                list.add(front);
                partition(res, list, s.substring(i));
                list.remove(list.size() - 1);
            }
        }
    }
    
    private boolean isPalindrome(String s) {
        if (s == null || s.length() == 0) return false;
        int lo = 0, hi = s.length() - 1;
        while (lo < hi) {
            if (s.charAt(lo) != s.charAt(hi)) return false;
            lo++;
            hi--;
        }
        return true;
    }
}

Reference：

http://stackoverflow.com/questions/24591616/whats-the-time-complexity-of-this-algorithm-for-palindrome-partitioning

http://blog.csdn.net/metasearch/article/details/4428865

https://en.wikipedia.org/wiki/Master_theorem

http://www.cnblogs.com/zhuli19901106/p/3570430.html

https://leetcode.com/discuss/18984/java-backtracking-solution

https://leetcode.com/discuss/9623/my-java-dp-only-solution-without-recursion-o-n-2

https://leetcode.com/discuss/41626/concise-java-solution

https://leetcode.com/discuss/4788/shouldnt-we-use-dp-in-addition-to-dfs