131. Palindrome Partitioning

简直开心，一遍通过，时间不是特别好，我觉得是isPalindrome用的是StirngBuilder方法不是数学方法的原因

不过很开心！！

 1     public List<List<String>> partition(String s) {
 2         List<List<String>> res = new ArrayList<List<String>>();
 3         if(s == null || s.length() == 0) {
 4             return res;
 5         }
 6         helper(res, s, 0, new ArrayList<String>());
 7         return res;
 8     }
 9     
10     private void helper(List<List<String>> res, String s, int start, List<String> curStr) {
11         int len = s.length();
12         if(start == len) {
13             res.add(new ArrayList<String>(curStr));
14             return;
15         }
16         for(int i = 1; i + start <= len; i++) {
17             String sub = s.substring(start, i + start);
18             if(isPalindrome(sub)) {
19                 curStr.add(sub);
20                 helper(res, s, start + i, curStr);
21                 curStr.remove(curStr.size() - 1);
22             }
23         }
24     }
25     
26     private boolean isPalindrome(String s) {
27         if(s == null || s.length() == 0) {
28             return true;
29         }
30         StringBuilder sb = new StringBuilder(s);
31         String rev = sb.reverse().toString();
32         return s.equals(rev);
33     }

 

posted @ 2016-06-22 06:46  warmland  阅读(119)  评论(0)  编辑  收藏  举报