leetcode 336. Palindrome Pairs

Given a list of unique words. Find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome.

Example 1:
Given words = ["bat", "tab", "cat"]
Return [[0, 1], [1, 0]]
The palindromes are ["battab", "tabbat"]

Example 2:
Given words = ["abcd", "dcba", "lls", "s", "sssll"]
Return [[0, 1], [1, 0], [3, 2], [2, 4]]
The palindromes are ["dcbaabcd", "abcddcba", "slls", "llssssll"]

思路：对每个单词w，以w打头，在w末尾加那些单词e可以使w+e成为回文，再看以w为结尾，加那些单词e可以使e+w成为回文，检索e是否在字典中。为快速检索，将字典装到hashmap里。stl里普通的map是基于rb-tree的，unodoered_map才是hash算法实现的，可以O(1)时间检索。

当前代码仍然可以优化的两个地方：

1.get_patch函数里不用遍历整个s的长度，也不用check，判断一下可以提前结束

2.答案判重多余

class Solution {
	bool check(string str)
	{
		int i = 0; int j = str.length() - 1;
		while (i <= j && str[i] == str[j])
		{
			i++; j--;
		}
		if (i > j)
			return true;
		return false;
	}
	string reverse(string s, int i, int j)//[i,j]闭区间，从0开始
	{
		string ss;
		for (int h = j; h >= i; h--)
			ss += s[h];
		return ss;
	}
	vector<string>get_patch(string &s)
	{
		vector<string>rr;
		rr.push_back(reverse(s, 0, s.length() - 1));
		rr.push_back(reverse(s, 0, s.length() - 2));
		int k = s.length() - 2;
		while (k >= 1)
		{
			if (k-1>=0&&s[k + 1] == s[k - 1])
			{
				if (check(s + reverse(s, 0, k -1- (s.length() - 1 - k))))
				rr.push_back(reverse(s, 0, k -1- (s.length() - 1 - k)));
			}
			if (s[k] == s[k + 1])
			{
				if (check(s + reverse(s, 0, k -1- (s.length() - 1 - k - 1))))
					rr.push_back(reverse(s, 0, k -1- (s.length() - 1 - k - 1)));
			}
			k--;
		}
		return rr;
	}
public:
	vector<vector<int>> palindromePairs(vector<string>& words) {
		set<vector<int>>re;
		unordered_map<string, int>hashword;
		bool flag = false;
		int index = -1;
		for (int i = 0; i < words.size(); i++)
		{
			hashword[words[i]] = i;
			if (words[i] == "")
			{
				flag = true;
				index = i;
			}
		}
		for (int i = 0; i < words.size(); i++)
		{
			if (flag&&check(words[i])&&i!=index)
			{
				vector<int>cc; cc.push_back(i); cc.push_back(index);
				re.insert(cc);
				cc.clear(); cc.push_back(index); cc.push_back(i);
				re.insert(cc);
			}
			vector<string>ss = get_patch(words[i]);
			for (int j = 0; j < ss.size();j++)
				if (hashword.find(ss[j]) != hashword.end() && hashword[ss[j]] != i)
				{
				vector<int>cc; cc.push_back(i); cc.push_back(hashword[ss[j]]);
				re.insert(cc);
				}
			string rs = reverse(words[i], 0, words[i].length() - 1);
			ss = get_patch(rs);
			for (int j = 0; j < ss.size(); j++)
			{
				string str = reverse(ss[j], 0, ss[j].length() - 1);
				if (hashword.find(str) != hashword.end() && hashword[str] != i)
				{
					vector<int>cc; cc.push_back(hashword[str]); cc.push_back(i);
					re.insert(cc);
				}
			}
		}
		return vector<vector<int>>(re.begin(),re.end());
	}
};

accept