131. Palindrome Partitioning-蒲公英云

131. Palindrome Partitioning

亦凉 2022-02-15 12:35 317阅读 0赞

Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]

本题又是一个回溯的题目。

public List<List<String>> partition(String s) {
        List<List<String>> ans = new ArrayList<>();
        if(s.length() == 0) return ans;
        backTracking(s, new ArrayList<>(), ans);
        return ans;
    }
    private void backTracking(String s, List<String> current, List<List<String>> ans) {
        if(s.length() == 0) {
            ans.add(new ArrayList<>(current));
            return;
        }
        for(int i = 0; i < s.length(); ++i) {
            String sub = s.substring(0, i + 1);
            if(isAPalindrome(sub)) {
                current.add(sub);
                backTracking(s.substring(i + 1), current, ans);
                current.remove(current.size() - 1);
            }
        }
    }
    private boolean isAPalindrome(String s) {
        for(int i = 0; i < s.length() / 2; ++i) {
            if(s.charAt(i) != s.charAt(s.length() - i - 1)) {
                return false;
            }
        }
        return true;
    }