leetcode解题: Palindrome Partitioning (131)

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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

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[
  ["aa","b"],
  ["a","a","b"]
]

解法1： O(n*2^n)

常规的backtracking解法，这里我们用一个变量cut来记录当前cut的位置， 然后从cut + 1 开始一个一个个试是否是palindrome。
复杂度的计算是由于一共有O(2^N)种可能的partition。
C++

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class Solution {
private:
    bool isPalindrome(string s) {
        for (int i = 0, j = s.size() - 1; i < j; ++i, --j) {
            if (s[i] != s[j]) {
                return false;
            }
        }
        return true;
    }
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        if (s.empty()) {
            return res;
        }
        vector<string> cur;
        helper(s, 0, cur, res);
        return res;
    }
    void helper(string s, int cut, vector<string>& cur, vector<vector<string>>& res) {
        if (cut == s.size()) {
            res.push_back(vector<string>(cur));
            return;
        }
        for (int i = cut + 1; i <= s.size(); ++i) {
            string prefix = s.substr(cut, i - cut);
            if (!isPalindrome(prefix)) {
                continue;
            }
            cur.push_back(prefix);
            helper(s, i, cur, res);
            cur.pop_back();
        }
    }
};

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class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return res;
        }
        helper(s, 0, new ArrayList<String>(), res);
        return res;
    }
    private void helper(String s, int pos, List<String> current, List<List<String>> res) {
        if (pos == s.length()) {
            res.add(new ArrayList<>(current));
            return;
        }
        for (int i = pos + 1; i <= s.length(); i++) {
            String cut = s.substring(pos, i);
            if (isPalindrome(cut)) {
                current.add(cut);
                helper(s, i, current, res);
                current.remove(current.size() - 1);
            }
        }
    }
    private boolean isPalindrome(String s) {
        if (s == null || s.length() == 0) {
            return true;
        }
        for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}