Palindrome Partitioning

https://leetcode.com/problems/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"]
  ]

解题思路：

这种题目一看就是DFS的题，中间用到一个 Valid Palindrome 的方法，判断一个string是不是palindrome。

对于每个位置start，取字符串的可能都是从start开始一直到结尾，所以要判断它是不是palindrome，是的话就加入current。对于下面的字符串继续DFS。到结尾了，回溯。

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> result = new LinkedList<List<String>>();
        if(s.length() == 0) {
            return result;
        }
        dfs(s, result, new LinkedList<String>(), 0);
        return result;
    }
    
    public void dfs(String s, List<List<String>> result, LinkedList<String> current, int start) {
        if(current.size() > 0 && !isPalindrome(current.getLast())) {
            return;
        }
        if(start == s.length()) {
            result.add(new LinkedList(current));
        }
        for(int i = 1; i < s.length() - start + 1; i++) {
            current.add(s.substring(start, start + i));
            dfs(s, result, current, start + i);
            current.remove(current.size() - 1);
        }
        
    }
    
    public boolean isPalindrome(String s) {
        if(s.length() == 0) {
            return true;
        }
        int start = 0, end = s.length() - 1;
        while(start < end) {
            if(s.charAt(start) != s.charAt(end)){
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
}

 //20180928

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<List<String>>();
        if (s == null || s.length() == 0) {
            return res;
        }
        List<String> cur = new ArrayList<String>();
        dfs(s, cur, res, 0, 0);
        return res;
    }
    
    public void dfs(String s, List<String> cur, List<List<String>> res, int start, int sumLength) {
        if (sumLength == s.length()) {
            res.add(new ArrayList<String>(cur));
            return;
        }
        
        for (int i = start; i < s.length(); i++) {
            String sub = s.substring(start, i + 1);
            if (!isPalindrome(sub)) {
                continue;
            }
            cur.add(sub);
            dfs(s, cur, res, i + 1, sumLength + sub.length());
            cur.remove(cur.size() - 1);
        }
    }
    
    public boolean isPalindrome(String s) {
        if (s == null || s.length() == 0 || s.length() == 1) {
            return true;
        }
        
        int left = 0, right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}