Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.
Example 1:
Input:"aabb"Output:["abba", "baab"]
Example 2:
Input:"abc"Output:[]
Hint:
- If a palindromic permutation exists, we just need to generate the first half of the string.
- To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.
这道题是之前那道
解法一： 下面这种方法和上面的方法很相似，不同之处来于求全排列的方法略有不同，上面那种方法是通过交换字符的位置来生成不同的字符串，而下面这种方法是通过加不同的字符来生成全排列字符串，参见代码如下： 解法二： 在来看一种利用了 std 提供的 next_permutation 函数来实现的方法，这样就大大减轻了我们的工作量，但是这种方法个人感觉算是有些投机取巧了，不知道面试的时候面试官允不允许这样做，贴上来拓宽一下思路也是好的： 解法三： Github 同步地址： https://github.com/grandyang/leetcode/issues/267 类似题目： 参考资料： https://leetcode.com/problems/palindrome-permutation-ii/ class Solution {
public:
vector<string> generatePalindromes(string s) {
unordered_set<string> res;
unordered_map<char, int> m;
string t = "", mid = "";
for (auto a : s) ++m[a];
for (auto it : m) {
if (it.second % 2 == 1) mid += it.first;
t += string(it.second / 2, it.first);
if (mid.size() > 1) return {};
}
permute(t, 0, mid, res);
return vector<string>(res.begin(), res.end());
}
void permute(string &t, int start, string mid, unordered_set<string> &res) {
if (start >= t.size()) {
res.insert(t + mid + string(t.rbegin(), t.rend()));
}
for (int i = start; i < t.size(); ++i) {
if (i != start && t[i] == t[start]) continue;
swap(t[i], t[start]);
permute(t, start + 1, mid, res);
swap(t[i], t[start]);
}
}
};
class Solution {
public:
vector<string> generatePalindromes(string s) {
vector<string> res;
unordered_map<char, int> m;
string t = "", mid = "";
for (auto a : s) ++m[a];
for (auto &it : m) {
if (it.second % 2 == 1) mid += it.first;
it.second /= 2;
t += string(it.second, it.first);
if (mid.size() > 1) return {};
}
permute(t, m, mid, "", res);
return res;
}
void permute(string &t, unordered_map<char, int> &m, string mid, string out, vector<string> &res) {
if (out.size() >= t.size()) {
res.push_back(out + mid + string(out.rbegin(), out.rend()));
return;
}
for (auto &it : m) {
if (it.second > 0) {
--it.second;
permute(t, m, mid, out + it.first, res);
++it.second;
}
}
}
};
class Solution {
public:
vector<string> generatePalindromes(string s) {
vector<string> res;
unordered_map<char, int> m;
string t = "", mid = "";
for (auto a : s) ++m[a];
for (auto it : m) {
if (it.second % 2 == 1) mid += it.first;
t += string(it.second / 2, it.first);
if (mid.size() > 1) return {};
}
sort(t.begin(), t.end());
do {
res.push_back(t + mid + string(t.rbegin(), t.rend()));
} while (next_permutation(t.begin(), t.end()));
return res;
}
};