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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"]
]
思路： 题目的要求是求所有的回文划分，因此一定是使用DFS（递归 + 回溯）模板。只需加入判断是否为回文的模块即可。
1 public class Solution { 2 /** 3 * @param s: A string 4 * @return: A list of lists of string 5 */ 6 public List<List<String>> partition(String s) { 7 if (s == null || s.length() == 0) { 8 return null; 9 } 10 List<List<String>> result = new ArrayList<>(); 11 List<String> path = new ArrayList<>(); 12 helper(s, result, path, 0); 13 return result; 14 } 15 private boolean isPalindrome(String s) { 16 int begin = 0; 17 int end = s.length() - 1; 18 while (begin < end) { 19 if (s.charAt(begin) != s.charAt(end)) { 20 return false; 21 } 22 ++begin; 23 --end; 24 } 25 return true; 26 } 27 private void helper(String s, List<List<String>> result, List<String> path, int pos) { 28 if (pos == s.length()) { 29 result.add(new ArrayList<String>(path)); 30 return; 31 } 32 for (int i = pos; i < s.length(); i++) { 33 String prefix = s.substring(pos, i + 1); 34 if (!isPalindrome(prefix)) { 35 continue; 36 } 37 path.add(prefix); 38 helper(s, result, path, i + 1); 39 path.remove(path.size() - 1); 40 } 41 } 42 }
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原文地址：http://www.cnblogs.com/FLAGyuri/p/5366698.html
