第十六周算法题

336. Palindrome Pairs

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• Total Accepted: 24269
• Total Submissions: 94125
• Difficulty: Hard
• Contributor: LeetCode

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"]

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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代码如下：

class Solution {public:    vector<vector<int>> palindromePairs(vector<string>& words) {        vector<vector<int>> res;        if (!words.size()) return res;        unordered_map<string, size_t> word_idx;        for (size_t i = 0; i < words.size(); ++i) {            word_idx[words[i]] = i;        }        vector<int> slu(2);        for (size_t i = 0; i < words.size(); ++i) {            size_t len = words[i].length();            for (size_t l = 0; l <= len; ++l) {                string left = words[i].substr(0, l);                string right = words[i].substr(l);                string rleft = left;                string rright = right;                reverse(rleft.begin(), rleft.end());                reverse(rright.begin(), rright.end());                if (word_idx.find(rleft) != word_idx.end()) {                    if (word_idx[rleft] != i && isPalindrome(right)) {                        slu[0] = i;                        slu[1] = word_idx[rleft];                        res.push_back(slu);                                            }                }                if (l != 0 && word_idx.find(rright) != word_idx.end()) {                    if (word_idx[rright] != i && isPalindrome(left)) {                        slu[0] = word_idx[rright];                        slu[1] = i;                        res.push_back(slu);                                            }                }            }        }        return res;    }        bool isPalindrome(string s) {        if (s.size() <= 1) return true;        size_t i = 0;         size_t j = s.size() - 1;        while (i < j) {            if (s[i++] != s[j--]) return false;        }                return true;    }};