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(Version 0.0)
这题的思路比较straightforward，可以看出会有overlapping的子问题，因此可以先按DP做法用一个二维数组存储对于某一substring是不是palindrome的判断，然后做backtracking时可以利用这个DP二维数组的结果来跳过重复的palindrome判断，不过如果不想使用DP做法其实每次都重复判断是否某一substring是palindrome也是可以过OJ的。用DP预处理的做法代码如下：
1 public class Solution { 2 public List<List<String>> partition(String s) { 3 List<List<String>> result = new ArrayList<>(); 4 boolean[][] isPalindrome = new boolean[s.length()][s.length()]; 5 for (int i = 0; i < isPalindrome.length; i++) { 6 isPalindrome[i][i] = true; 7 } 8 for (int l = 2; l <= s.length(); l++) { 9 for (int i = 0; i + l <= s.length(); i++) { 10 int j = i + l - 1; 11 isPalindrome[i][j] = (isPalindrome[i + 1][j - 1] || l == 2) && s.charAt(i) == s.charAt(j); 12 } 13 } 14 List<String> list = new ArrayList<>(); 15 helper(result, list, s, isPalindrome, 0); 16 return result; 17 } 18 19 private void helper(List<List<String>> result, List<String> list, String s, boolean[][] isPalindrome, int start) { 20 if (start == s.length()) { 21 List<String> temp = new ArrayList<>(list); 22 result.add(temp); 23 return; 24 } 25 for (int end = start; end < s.length(); end++) { 26 if (isPalindrome[start][end]) { 27 list.add(s.substring(start, end + 1)); 28 helper(result, list, s, isPalindrome, end + 1); 29 list.remove(list.size() - 1); 30 } 31 } 32 } 33 }
[LeetCode] Palindrome Partitioning
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原文地址：http://www.cnblogs.com/icecreamdeqinw/p/4331680.html
