【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);
}
}
}
}