leetcode:Palindrome Partitioning

最新推荐文章于 2019-07-18 15:18:00 发布

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本文介绍了一种用于找出所有可能的回文字符串分割方案的算法。该算法通过递归方式实现，利用动态规划预先计算子串是否为回文，从而高效地解决此问题。

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"]
  ]

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class Solution {
    
private:
    void partitionHelper(string &s, int idx, vector<vector<bool>> &dp, vector<string> &curVtr, vector<vector<string>> &retVtr) {
        
        if (idx == s.size())
        {
            retVtr.push_back(curVtr);
            return;
        }
            
            
        for (int len=1; len<=(int)s.size()-idx; len++)
        {
            if (dp[idx][idx+len-1])
            {
                curVtr.push_back(s.substr(idx, len));
                partitionHelper(s, idx+len, dp, curVtr, retVtr);
                curVtr.pop_back();
            }
        }
    }
    
public:
    vector<vector<string>> partition(string s) {
        
        vector<vector<string>> retVtr;
        vector<string> curVtr;
        
        vector<vector<bool>> dp(s.size(), vector<bool>(s.size()));
        for (int i=0; i<s.size(); i++)  
            dp[i][i] = true;  
              
        for (int i=1; i<s.size(); i++)  
            if (s[i-1] == s[i])  
                dp[i-1][i] = true;  
                  
        for (int len=2; len<=s.size()-1; len++)  
        {  
            //这里等号要注意  
            for (int i=0; i<=(int)s.size()-1-len; i++)  
            {  
                if (s[i] == s[i+len] && dp[i+1][i+len-1])  
                    dp[i][i+len] = true;  
            }  
        }  
        
        partitionHelper(s, 0, dp, curVtr, retVtr);
        
        return retVtr;
    }
};