Java for LeetCode 131 Palindrome Partitioning

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"]
  ]
解题思路一：

将s分为左右两个部分，分别求出两边的partition，然后粘一块即可，JAVA实现如下：

	static public List<List<String>> partition(String s) {
		Set<List<String>> set = new HashSet<List<String>>();
		if (isPalindrome(s)) {
			List<String> alist = new ArrayList<String>();
			alist.add(s);
			set.add(alist);
		}
		for (int i = 1; i < s.length(); i++) {
			List<List<String>> left = partition(s.substring(0, i));
			List<List<String>> right = partition(s.substring(i, s.length()));
			for (List<String> aLeft : left)
				for (List<String> aRight : right) {
					List<String> alist = new ArrayList<String>(aLeft);
					alist.addAll(aRight);
					set.add(new ArrayList<String>(alist));
				}
		}
		return new ArrayList<List<String>>(set);
	}

	static boolean isPalindrome(String s) {
		int left = 0;
		int right = s.length() - 1;
		while (left < right)
			if (s.charAt(left++) != s.charAt(right--))
				return false;
		return true;
	}

结果TLE

解题思路二：

修改下思路一，从左边入手，如果左边是Palindrome，对右边求一个partition，这样求得的结果也不会重复，这样就可以AC了，JAVA实现如下：

	static public List<List<String>> partition(String s) {
		ArrayList<List<String>> list = new ArrayList<List<String>>();
		if (isPalindrome(s)) {
			List<String> alist = new ArrayList<String>();
			alist.add(s);
			list.add(alist);
		}
		for (int i = 1; i < s.length(); i++)
			if (isPalindrome(s.substring(0, i))) {
				List<String> aLeft = new ArrayList<String>();
				aLeft.add(s.substring(0, i));
				List<List<String>> right = partition(s.substring(i, s.length()));
				for (List<String> aRight : right) {
					List<String> alist = new ArrayList<String>(aLeft);
					alist.addAll(aRight);
					list.add(new ArrayList<String>(alist));
				}
			}
		return list;
	}

	static boolean isPalindrome(String s) {
		int left = 0;
		int right = s.length() - 1;
		while (left < right)
			if (s.charAt(left++) != s.charAt(right--))
				return false;
		return true;
	}