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.

Example:

Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]

题目大意：给定一个字符串s，对其进行划分，是的划分后的所有字串都是回文串， 返回所有可能的划分。

思路：回溯法

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string> > res;
        vector<string> temp;
        backtrack(s, 0, temp, res);
        return res;
    }
private:
    bool isPalindrome(string s, int i, int j) {
        while (i < j) {
            if (s[i] != s[j])
                return false;
            i++;
            j--;
        }
        return true;
    }
    void backtrack(string &s, int i, vector<string> &v, vector<vector<string>> &res) {
        if ((i == s.length()) && !v.empty()) {
            res.push_back(v);
            return ;
        }
        for (int k = i; k < s.length(); ++k) {
            if (isPalindrome(s, i, k)) {
                v.push_back(s.substr(i, k - i + 1));
                backtrack(s, k + 1, v, res);
                v.pop_back();
            }
        }
    }
};

时间复杂度：

T(n) = n * [T(n - 1) + ... + T(1)]

T(n + 1) = (n + 1） * [ T(n) + T(n - 1) + ... + T(1)] = (n + 1) * [ T(n) + 1/n * T(n)]

==> T(n) = n * 2^n

优化：判断回文串的时候有大量重复。

class Solution {
public:
    vector<vector<string>> partition(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n, 0)); //dp[i][j]表示下表i到下表j的字串是否为回文串
        for (int i = 0; i < n; i++) {
            dp[i][i] = 1;
        }
        
        for (int i = 0; i < n - 1; i++) {
            if (s[i] == s[i+1]) {
                dp[i][i+1] = 1;
            }
        }
        
        for (int len = 2; len < n; len++) {
            for (int i = 0; i < n; i++) {
                if (i + len >= n) {
                    break;
                } else {
                    if (s[i] == s[i + len]) {
                        dp[i][i+len] = dp[i+1][i+len-1] || dp[i][i+len];
                    }
                }
            }
        }
        vector<vector<string> > res;
        vector<string> temp;
        backtrack(s, 0, temp, res, dp);
        return res;
    }
private:
    void backtrack(string &s, int i, vector<string> &v, vector<vector<string>> &res, vector<vector<int>> &dp) {
        if ((i == s.length()) && !v.empty()) {
            res.push_back(v);
            return ;
        }
        for (int k = i; k < s.length(); ++k) {
            if (dp[i][k]) {
                v.push_back(s.substr(i, k - i + 1));
                backtrack(s, k + 1, v, res, dp);
                v.pop_back();
            }
        }
    }
};