Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
public class Solution { /** * @param s: A string * @return: A list of lists of string */ public List<List<String>> partition(String s) { List<List<String>> results = new ArrayList<>(); if (s == null || s == "") { return results; } ArrayList<String> result = new ArrayList<>(); dfsHelper(s, 0, results, result); return results; } PRivate void dfsHelper(String s, int start, List<List<String>> results, ArrayList<String> result) { if (start == s.length()) { results.add(new ArrayList<String>(result)); } //判断清楚是对i、还是index进行递归！！！ for (int i = start; i < s.length(); i++) { String subString = s.substring(start, i + 1); if (!isPalindrome(subString)) { continue; } result.add(subString); dfsHelper(s, i + 1, results, result); result.remove(result.size() - 1); } } private boolean isPalindrome(String s) { int start = 0; int end = s.length() - 1; while (start < end) { if (s.charAt(start) != s.charAt(end)) { return false; } start++; end--; } return true; }}新闻热点
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