79 lines
2.6 KiB
C++
79 lines
2.6 KiB
C++
// Source : https://leetcode.com/problems/palindrome-pairs/
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// Author : Hao Chen
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// Date : 2017-03-22
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/***************************************************************************************
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*
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* Given a list of unique words, find all pairs of distinct indices (i, j) in the given
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* list, so that the concatenation of the two words, i.e. words[i] + words[j] is a
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* palindrome.
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*
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* Example 1:
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* Given words = ["bat", "tab", "cat"]
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* Return [[0, 1], [1, 0]]
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* The palindromes are ["battab", "tabbat"]
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*
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* Example 2:
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* Given words = ["abcd", "dcba", "lls", "s", "sssll"]
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* Return [[0, 1], [1, 0], [3, 2], [2, 4]]
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* The palindromes are ["dcbaabcd", "abcddcba", "slls", "llssssll"]
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*
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* Credits:Special thanks to @dietpepsi for adding this problem and creating all test
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* cases.
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*
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***************************************************************************************/
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class Solution {
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public:
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bool isPalindrome(string& str) {
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int left = 0, right = str.size() - 1;
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while( left < right) {
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if (str[left++] != str[right--]) return false;
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}
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return true;
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}
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vector<vector<int>> palindromePairs(vector<string>& words) {
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unordered_map<string, int> dict;
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for(int i=0; i<words.size(); i++){
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string w = words[i];
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reverse(w.begin(), w.end());
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dict[w] = i;
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}
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vector<vector<int>> result;
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//egde case: deal with empty string
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if ( dict.find("") != dict.end() ) {
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for(int i=0; i<words.size(); i++) {
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if ( isPalindrome(words[i]) && dict[""] != i ) {
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result.push_back( { dict[""], i } );
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}
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}
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}
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for(int i=0; i<words.size(); i++) {
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for (int j=0; j<words[i].size(); j++) {
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//split the word to 2 parts
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string left = words[i].substr(0, j);
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string right = words[i].substr(j, words[i].size() - j);
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// if the `left` is found, which means there is a words has reversed sequence with it.
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// then we can check the `right` part is Palindrome or not.
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if ( dict.find(left) != dict.end() && isPalindrome(right) && dict[left] != i ) {
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result.push_back( { i, dict[left] } );
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}
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if (dict.find(right) != dict.end() && isPalindrome(left) && dict[right] != i ) {
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result.push_back( { dict[right], i } );
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}
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}
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}
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return result;
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}
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};
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