leetcode.336. Palindrome Pairs

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"]

Steps:

Traverse the array, build map. Key is the reversed string, value is index in array (0 based)

Edge case - check if empty string exists. It’s interesting that for given words {“a”, “”}, it’s expected to return two results [0,1] and [1,0]. Since my main logic can cover [0, 1] concatenate(“a”, “”), so as to cover the other situation concatenate(“”,
“a”), I need to traverse the words array again, find the palindrome word candidate except “” itself, and add pair(“”, palindrome word) to the final answer.

Main logic part. Partition the word into left and right, and see 1) if there exists a candidate in map equals the left side of current word, and right side of current word is palindrome, so concatenate(current word, candidate) forms a pair:
left | right | candidate. 2) same for checking the right side of current word:
candidate | left | right.

时间复杂度为： O(n * k^2)， 其中n为单词的数量，k为单词的位数。

class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> ret;
unordered_map<string, int> dict;

for (int i = 0; i < words.size(); ++i) {
string s = words[i];
reverse(s.begin(), s.end());
dict[s] = i;
}

for (int i = 0; i < words.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 (left.empty()) {
ret.push_back({dict[left], i});
}
}
if (dict.find(right) != dict.end() && IsPalindrome(left) && dict[right] != i) {
ret.push_back({dict[right], i});
}

}
}
return ret;

}
bool IsPalindrome(const string &s) {
for (int left = 0, right = s.size() - 1; left < right; ++left, --right) {
if (s[left] != s[right]) return false;
}
return true;
}

};