336. Palindrome Pairs

https://leetcode.com/problems/palindrome-pairs/description/

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"]

Thoughts

在一堆字符串中找出能拼成回文的pair。一个str可以看成l和r两部分，然后可能有两种和其它str拼成回文的方法：rev_r l r和l r rev_l。因此把组内所有reverserd存进hash map中，然后对每一个词遍历所有的划分，看l自身是回文的情况下，map中有没有rev_r；r同理。因为空串的存在，所有的划分包含了null(l) s(r)和s(l) null。 为了避免rev_r null r和l null rev_l 给出重复结果，第二个判断条件加检查l是否为空。

Code

/*
 * @lc app=leetcode id=336 lang=cpp
 *
 * [336] Palindrome Pairs
 */

// @lc code=start
class Solution {
public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
        const int N = words.size();
        unordered_map<string, int> dict;
        for (int i = 0; i < N; ++i) {
            auto rw = words[i];
            reverse(rw.begin(), rw.end());
            dict[rw] = i;
        }
        const auto palind = [](const string &s) {
            for (int i = 0, j = s.length() - 1; i < j; ++i, --j) {
                if (s[i] != s[j]) return false;
            }
            return true;
        };
        vector<vector<int>> res;
        for (int i = 0; i < N; ++i) {
            const auto w = words[i];
            for (int j = 0; j <= w.length(); ++j) {
                const auto l = w.substr(0, j), r = w.substr(j);
                if (dict.count(l) && dict[l] != i && palind(r)) {
                    res.push_back({i, dict[l]});
                }
                if (dict.count(r) && dict[r] != i && palind(l) && !l.empty()) {
                    res.push_back({dict[r], i});
                }
            }
        }
        return res;
    }
};
// @lc code=end

Analysis

时间复杂度O(N).