Palindrome Partitioning 分割字符串为回文@LeetCode

DFS回溯又一经典应用

package Level4;

import java.util.ArrayList;
import java.util.Hashtable;

/**
 *  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"]
  ]
 *
 */
public class S136 {

	public static void main(String[] args) {
		String s = "aab";
		System.out.println(partition(s));
	}
	
	public static ArrayList<ArrayList<String>> partition(String s) {
		ArrayList<ArrayList<String>> ret = new ArrayList<ArrayList<String>>();
		ArrayList<String> al = new ArrayList<String>();
		Hashtable<String, Boolean> ht = new Hashtable<String, Boolean>();
		dfs(s, 0, ht, ret, al);
		return ret;
    }
	
	public static void dfs(String s, int start, Hashtable<String, Boolean> ht, ArrayList<ArrayList<String>> ret, ArrayList<String> al){
		if(start == s.length()){			// 返回条件
			ret.add(new ArrayList<String>(al));
			return;
		}
		
		for(int i=start; i<s.length(); i++){				// 拓展状态
			boolean isPalindrome = false;
			String substr = s.substring(start, i+1);		// 子字符串
			if(ht.get(substr) != null){				// 先尝试在hashtable中查找，避免重复查找
				isPalindrome = ht.get(substr);
			}else{										// 判断是否palindrome，放入ht
				isPalindrome = checkPalindrome(substr);
				ht.put(substr, isPalindrome);
			}
			if(isPalindrome){		
				al.add(substr);				// 执行拓展动作
				dfs(s, i+1, ht, ret, al);
				al.remove(al.size()-1);	// 撤销动作
			}
		}
	}
	
	public static boolean checkPalindrome(String s){
		if(s.length() == 0){
			return true;
		}
		for(int i=0; i<=s.length()/2; i++){
			if(s.charAt(i) != s.charAt(s.length()-1-i)){
				return false;
			}
		}
		return true;
	}

}

不用Hashtable也可以通过

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> ret = new ArrayList<List<String>>();
        rec(ret, new ArrayList<String>(), s, 0);
        return ret;
    }
    
    public void rec(List<List<String>> ret, List<String> list, String s, int pos) {
        if(pos == s.length()) {
            ret.add(new ArrayList<String>(list));
            return;
        }
        
        for(int i=pos; i<s.length(); i++) {
            String substr = s.substring(pos, i+1);
            if(isPalindrome(substr)) {
                list.add(substr);
                rec(ret, list, s, i+1);
                list.remove(list.size()-1);
            } 
        }
    }
    
    public boolean isPalindrome(String s) {
        int left = 0, right = s.length()-1;
        while(left <= right) {
            if(s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}