Palindrome Pairs
Desicription
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:
代码语言：javascript复制Input: ["abcd","dcba","lls","s","sssll"]
Output: [[0,1],[1,0],[3,2],[2,4]]
Explanation: The palindromes are ["dcbaabcd","abcddcba","slls","llssssll"]Example 2:
代码语言：javascript复制Input: ["bat","tab","cat"]
Output: [[0,1],[1,0]]
Explanation: The palindromes are ["battab","tabbat"]Solution
代码语言：javascript复制class Solution {
public:
std::vector<std::vector<int>> palindromePairs(std::vector<std::string>& words) {
auto res = std::vector<std::vector<int>>();
auto indexMap = std::map<std::string, int>();
auto sizeSet = std::set<int>();
for(int i = 0; i < words.size(); i ) {
indexMap.insert({words[i], i});
sizeSet.insert(words[i].size());
}
for(int i = 0; i < words.size(); i ) {
auto word = words[i];
std::reverse(word.begin(), word.end());
if(indexMap.find(word) != indexMap.end() && indexMap[word] != i) {
res.push_back({i, indexMap[word]});
}
auto rightIterator = sizeSet.find(word.size());
for(auto it = sizeSet.begin(); it != rightIterator; it ) {
if(isPalindrome(word, 0, word.size() - *it - 1) && indexMap.find(word.substr(word.size() - *it)) != indexMap.end()) {
res.push_back({i, indexMap[word.substr(word.size() - *it)]});
}
if(isPalindrome(word, *it, word.size() - 1) && indexMap.find(word.substr(0, *it)) != indexMap.end()) {
res.push_back({indexMap[word.substr(0, *it)], i});
}
}
}
return res;
}
private:
static bool isPalindrome(const std::string &word, int left, int right) {
while(left < right) {
if(word[left ] != word[right--]) {
return false;
}
}
return true;
}
};
