Palindrome Pairs

嘗試着理解了一下Trie樹，原理解釋見：https://discuss.leetcode.com/topic/39585/o-n-k-2-java-solution-with-trie-structure-n-total-number-of-words-k-average-length-of-each-word

我把代碼改成了C++，使用了shared_ptr，做了簡單的註釋，已經AC

class TrieNode {
public:
	vector<shared_ptr<TrieNode>> next;
	int index=-1;//截止到當前節點爲止是否構成words中的一個字符串，並記錄字符串的索引位置
	vector<int> list;//截止到當前節點爲止形成了一個字符串A（可能是words中的元素，也可能不是），list裏保存words中所有含有字符串A的元素的位置
	TrieNode() { next.resize(26); }
};
class Solution {
public:
	vector<vector<int>> palindromePairs(vector<string>& words) {
		vector<vector<int>> res;

		shared_ptr<TrieNode> root(new TrieNode());
		for (int i = 0; i < words.size(); i++)
			addWord(root, words[i], i);
		for (int i = 0; i < words.size(); i++)
			search(words, i, root, res);

		return res;
	}

	void addWord(shared_ptr<TrieNode> root, string& word, int index) {
		for (int i = word.size() - 1; i >= 0; i--) {
			int j = word[i] - 'a';
			if (root->next[j] == NULL) 
				root->next[j] = make_shared<TrieNode>();
			if (isPalindrome(word, 0, i)) 
				root->list.push_back(index);
			root = root->next[j];
		}

		root->list.push_back(index);
		root->index = index;
	}

	void search(vector<string>& words, int i, shared_ptr<TrieNode> root, vector<vector<int>>& res) {
		for (int j = 0; j < words[i].length(); j++) {
			//下面的這個if用來處理“ac”，“cabb”這類情況，這裏就體現出TrieNode類的設計的tricky之處了
			if (root->index >= 0 && root->index != i && isPalindrome(words[i], j, words[i].length() - 1)) {
				res.push_back(vector<int>{i, root->index});
			}

			root = root->next[words[i][j] - 'a'];
			if (root == NULL) return;
		}
		//下面這個循環用來處理“aaa”，“aaa”這類情況
		for (int j : root->list) {
			if (i == j) continue;
			res.push_back(vector<int>{i, j});
		}
	}

	bool isPalindrome(string& word, int i, int j) {
		while (i < j) {
			if (word[i++] != word[j--]) return false;
		}

		return true;
	}
};