# [LeetCode] Palindrome Pairs 回文对

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.

class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> res;
unordered_map<string, int> m;
set<int> s;
for (int i = 0; i < words.size(); ++i) {
m[words[i]] = i;
s.insert(words[i].size());
}
for (int i = 0; i < words.size(); ++i) {
string t = words[i];
int len = t.size();
reverse(t.begin(), t.end());
if (m.count(t) && m[t] != i) {
res.push_back({i, m[t]});
}
auto a = s.find(len);
for (auto it = s.begin(); it != a; ++it) {
int d = *it;
if (isValid(t, 0, len - d - 1) && m.count(t.substr(len - d))) {
res.push_back({i, m[t.substr(len - d)]});
}
if (isValid(t, d, len - 1) && m.count(t.substr(0, d))) {
res.push_back({m[t.substr(0, d)], i});
}
}
}
return res;
}
bool isValid(string t, int left, int right) {
while (left < right) {
if (t[left++] != t[right--]) return false;
}
return true;
}
};

class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> res;
unordered_map<string, int> m;
for (int i = 0; i < words.size(); ++i) m[words[i]] = i;
for (int i = 0; i < words.size(); ++i) {
int l = 0, r = 0;
while (l <= r) {
string t = words[i].substr(l, r - l);
reverse(t.begin(), t.end());
if (m.count(t) && i != m[t] && isValid(words[i].substr(l == 0 ? r : 0, l == 0 ? words[i].size() - r: l))) {
if (l == 0) res.push_back({i, m[t]});
else res.push_back({m[t], i});
}
if (r < words[i].size()) ++r;
else ++l;
}
}
return res;
}
bool isValid(string t) {
for (int i = 0; i < t.size() / 2; ++i) {
if (t[i] != t[t.size() - 1 - i]) return false;
}
return true;
}
};

https://leetcode.com/discuss/91562/my-c-solution-275ms-worst-case-o-n-2

https://leetcode.com/discuss/91531/accepted-short-java-solution-using-hashmap

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-03-13 15:00  Grandyang  阅读(22531)  评论(0编辑  收藏  举报