Palindrome Pairs

题目描述：

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"]

解题思路：

对于一个字符串，例如"abcd"，其反转后的字符串为"dcba"，把反转后的字符串加到原字符串的前面和后面都可以构成一个回文字符串，"dcbaabcd"和"abcddcba"均为回文字符串。因此可以对words中的每一个非空字符串求其反转后的字符串，然后判断反转后的字符串是否也在words中即可。由于把反转后的字符串的一部分加到原字符串就可以构成回文字符串，例如对于字符串"sssll"，其反转后的字符串为"llsss"，把"llsss"的子串"lls"加到原字符串就可以构成回文字符串，因此需要对反转后的字符串的每一个子串都进行测试，即把反转后的字符串的每一个子串加到原字符串，看是否可以构成回文字符串。

AC代码如下：

class Solution {
public:vector<vector<int>> palindromePairs(vector<string>& words) {set<vector<int>> ans;int n = words.size();if (n <= 1) return vector<vector<int>>();map<string, int> dicts;for (int i = 0; i < n; ++i){dicts[words[i]] = i;}for (int i = 0; i < n; ++i){if (words[i] == "") continue;if (isPalindrome(words[i])){if (dicts.find("") != dicts.end()){vector<int> pos1;pos1.push_back(dicts[""]);pos1.push_back(i);ans.insert(pos1);vector<int> pos2;pos2.push_back(i);pos2.push_back(dicts[""]);ans.insert(pos2);}}string str_reverse = string(words[i].rbegin(), words[i].rend());int len = str_reverse.size();for (int j = 1; j <= len; ++j){string str1 = str_reverse.substr(0, j);//forwardif (isPalindrome(str1 + words[i]) && dicts.find(str1) != dicts.end()){vector<int> pos;pos.push_back(dicts[str1]);pos.push_back(i);if (dicts[str1]!=i)ans.insert(pos);}//backwardstring str2 = str_reverse.substr(len - j);if (isPalindrome(words[i] + str2) && dicts.find(str2) != dicts.end()){vector<int> pos;pos.push_back(i);pos.push_back(dicts[str2]);if (dicts[str2] != i)ans.insert(pos);}}}return vector<vector<int>>(ans.begin(), ans.end());}
private:bool isPalindrome(const string& str){for (int i = 0, j = str.size() - 1; i < j; ++i, --j)if (str[i] != str[j]) return false;return true;}
};

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