LeetCodePalindrome Partitioning 解题报告

Posted 2020-09-20 blfbuaa

【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"]
  ]

【回溯】

public class Solution {
    public ArrayList<ArrayList<String>> partition(String s) {
    	ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
        ArrayList<String> list = new ArrayList<String>();
        
    	if (s == null || s.length() == 0)
    		return result;
        
        calResult(result,list,s);
        return result;
    }
    
    /**
     * 推断一个字符串是否是回文字符串
     */
    private boolean isPalindrome(String str){
    	
    	int i = 0;
    	int j = str.length() - 1;
    	while (i < j){
    		if (str.charAt(i) != str.charAt(j)){
    			return false;
    		}
    		i++;
    		j--;
    	}
    	return true;
    }
    
    /**
     * 回溯
     * @param result : 终于要的结果集 ArrayList<ArrayList<String>>
     * @param list : 当前已经增加的集合 ArrayList<String>
     * @param str : 当前要处理的字符串
     */
    private void calResult(ArrayList<ArrayList<String>> result
    		, ArrayList<String> list
    		, String str)
    {
    	//当处理到传入的字符串长度等于0,则这个集合list满足条件，增加到结果集中
    	if (str.length() == 0)
    		result.add(new ArrayList<String>(list));
    	int len = str.length();
    	//递归调用
    	//字符串由前往后，先推断str.substring(0, i)是否是回文字符串
    	//假设是的话，继续调用函数calResult，把str.substring(i)字符串传入做处理
    	for (int i=1; i<=len; ++i){
    		String subStr = str.substring(0, i);
    		if (isPalindrome(subStr)){
    			list.add(subStr);
    			String restSubStr = str.substring(i);
    			calResult(result,list,restSubStr);
    			list.remove(list.size()-1);
    		}
    	}
    }
}