LeetCode: Palindrome Partitioning 解题报告

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"]
  ]

Solution 0:

直接用DFS 做，实际上也是可以过Leetcode的检查的。

 1 public List<List<String>> partition(String s) {
 2         List<List<String>> ret = new ArrayList<List<String>>();
 3         if (s == null) {
 4             return ret;
 5         }
 6         
 7         dfs(s, 0, new ArrayList<String>(), ret);
 8         return ret;
 9     }
10     
11     public static void dfs(String s, int index, List<String> path, List<List<String>> ret) {
12         int len = s.length();
13         if (index == len) {
14             ret.add(new ArrayList<String>(path));
15             return;
16         }
17         
18         for (int i = index; i < len; i++) {
19             String sub = s.substring(index, i + 1);
20             if (!isPalindrome(sub)) {
21                 continue;
22             }
23             
24             path.add(sub);
25             dfs(s, i + 1, path, ret);
26             path.remove(path.size() - 1);
27         }
28     }
29     
30     public static boolean isPalindrome(String s) {
31         int len = s.length();
32         int left = 0;
33         int right = len - 1;
34         
35         while (left < right) {
36             if (s.charAt(left) != s.charAt(right)) {
37                 return false;
38             }
39             left++;
40             right--;
41         }
42         
43         return true;
44     }

View Code