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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"]
class Solution { private: bool isPalindrome(string s) { int left = 0; int right = s.length() - 1; while (left < right) { if (s[left++] != s[right--]) return false; } return true; } public: vector<vector<int>> palindromePairs(vector<string>& words) { vector<vector<int>> ret; unordered_map<string, int> dict; int size = words.size(); for (int i = 0; i < size; i++) { string key = words[i]; reverse(key.begin(), key.end()); dict[key] = i; } if (dict.find("") != dict.end()) { for (int i = 0; i < size; i++) { if (i == dict[""]) continue; if (isPalindrome(words[i])) ret.push_back( { dict[""], i }); } } for (int i = 0; i < size; i++) { for (int j = 0; j < words[i].size(); j++) { string left = words[i].substr(0, j); string right = words[i].substr(j, words[i].size() - j); if (dict.find(left) != dict.end() && isPalindrome(right) && dict[left] != i) { ret.push_back( { i, dict[left] }); } if (dict.find(right) != dict.end() && isPalindrome(left) && dict[right] != i) { ret.push_back( { dict[right], i }); } } } return ret; } };
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原文地址:http://www.cnblogs.com/wxquare/p/5864088.html