[LeetCode]Palindrome Partitioning

最新推荐文章于 2019-09-02 10:48:14 发布

原创最新推荐文章于 2019-09-02 10:48:14 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

152 篇文章

订阅专栏

本文介绍了一种用于将字符串分割成多个回文子串的有效算法。通过递归深度优先搜索(DFS)和动态规划的方法，该算法能够找出所有可能的分割方案。文章详细解释了核心代码实现，并提供了两种不同的实现思路。

class Solution {
public:
	bool IsPalindrome(string str, int startIdx, int endIdx)
	{
		while (startIdx < endIdx)
		{
			if(str[startIdx] != str[endIdx])
				return false;
			startIdx++;
			endIdx--;
		}
		return true;
	}
	void DFS(string& str, int startIdx, vector<string>& curSubStrs, vector<vector<string>>& result)
	{
		if (startIdx == str.size())
		{
			result.push_back(curSubStrs);
			return;
		}
		for (int i = startIdx; i < str.size(); ++i)
		{
			if ( IsPalindrome(str, startIdx, i) )
			{
				curSubStrs.push_back(str.substr(startIdx, i-startIdx+1));
				DFS(str, i+1, curSubStrs, result);
				curSubStrs.pop_back();
			}
		}
	}
	vector<vector<string>> partition(string s) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		vector<string> curSubStrs;
		vector<vector<string>> result;
		DFS(s, 0, curSubStrs, result);
		return result;
	}
};

second time

class Solution {
public:
    void partitionUtil(string& s, vector<vector<bool> >& isP, int curIdx, vector<int>& curPath, vector<vector<string> >& allPath)
    {
        if(curIdx == s.size())
        {
            vector<string> curStr;
            //
            int end = s.size();
            for(int i = curPath.size()-1; i >= 0; --i)
            {
                curStr.push_back(s.substr(curPath[i], end-curPath[i]));
                end = curPath[i];
            }
            reverse(curStr.begin(), curStr.end());
            allPath.push_back(curStr);
            return;
        }
        
        for(int i = curIdx; i < s.size(); ++i)
        {
            if(isP[curIdx][i])
            {
                curPath.push_back(curIdx);
                partitionUtil(s, isP, i+1, curPath, allPath);
                curPath.pop_back();
            }
        }
    }
    vector<vector<string>> partition(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = s.size();
        vector<vector<bool> > isP(n, vector<bool>(n, false));
        for(int i = n-1; i >= 0; --i)
        {
            for(int j = i; j < n; ++j)
            {
                if(s[i] == s[j]) isP[i][j] = (j-1 <= i+1) || isP[i+1][j-1];
            }
        }
        
        vector<vector<string> > allPath;
        vector<int> curPath;
        partitionUtil(s, isP, 0, curPath, allPath);
        return allPath;
    }
};