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_131_PalindromePartitioning.java
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_131_PalindromePartitioning.java
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package leetcode_1To300;
import java.util.ArrayList;
import java.util.List;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _131_PalindromePartitioning {
/**
* 131. 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"]
]
time: O(2^n) space O(n)
* @param s
* @return
*/
public List<List<String>> partition(String s) {
List<List<String>> res = new ArrayList<>();
if (s == null || s.length() == 0) return res;
helper(res, new ArrayList<>(), s);
return res;
}
public void helper(List<List<String>> res, List<String> list, String s) {
if (s.length() == 0) {
res.add(new ArrayList<>(list));
return;
}
for (int i = 0; i < s.length(); i++) {
if (isPalindrome(s.substring(0, i + 1))) {
list.add(s.substring(0, i + 1));
helper(res, list, s.substring(i + 1));
list.remove(list.size() - 1);
}
}
}
public boolean isPalindrome(String s) {
for (int i = 0; i < s.length() / 2; i++) {
if (s.charAt(i) != s.charAt(s.length() - i - 1)) {
return false;
}
}
return true;
}
}