Palindrome Permutation I & II

Palindrome Permutation I

Given a string, determine if a permutation of the string could form a palindrome.

For example,
"code" -> False, "aab" -> True, "carerac" -> True.

Hint:

1. Consider the palindromes of odd vs even length. What difference do you notice?
2. Count the frequency of each character.
3. If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times

分析：

　　这个问题不需要判断是否是回文字符串，而是判断是否能组成回文字符串，换句话说就是字母在原字符串中的顺序无关。

解法：

　　可根据回文定义得出，即允许出现奇数次的字母种数最多为1

证明：

　　充分性，将出现奇数次的字母放在中间，若无出现奇数次的字母，则直接做下一步，然后从中间向两边依次放置出现偶数次的字母，满足；

　　必要性，任意回文字符串都满足中轴对称，偶数个字母则有出现奇数次的字母种数为0，奇数个字母则有出现奇数次的字母种数为1，满足；

代码：

bool isPermutation(string str){
    vector<char> bin(26 ,0);
    for(char c : str)
        bin[int(c - ‘a‘)] ^= 1;
    int count = 0;
    for(int i : bin)
        count += i;
    return count <= 1;
}

Palindrome Permutation II

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

For example:

Given s = "aabb", return ["abba", "baab"].

Given s = "abc", return [].

Hint:

1. If a palindromic permutation exists, we just need to generate the first half of the string.
2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

分析：

　　这个问题相比上个问题，是个后续输出工作，直接排列所有情况即可，证明比较直观。

解法：

　　直接排列。小技巧同Hint. 1给出的，只需要得出一边的排列。

代码：

void dfs(unordered_set<string> &uset, string str, vector<int> bin, int total) {
    if(total == 0) {
        uset.insert(str);
        return;
    }
    for(int i = 0; i < bin.size(); i++) {
        if(bin[i] == 0)
            continue;
        bin[i]--;
        dfs(uset, str + char(i + ‘a‘), bin, total - 1);
        bin[i]++;
    }
    return;
}
vector<string> permutation(string str){
    vector<int> bin(26 ,0);
    for(char c : str)
        bin[int(c - ‘a‘)]++;
    int count = 0, total = 0;
    char record;
    for(int i = 0; i < bin.size(); i++) {
        total += bin[i];
        if((bin[i] & 1) == 1) {
            record = char(i + ‘a‘);
            count++;
        }
    }
    vector<string> vs;
    if(count > 1)
        return vs;
    for(int &i : bin)
        i /= 2;
    unordered_set<string> uset;
    dfs(uset, "", bin, total / 2);
    for(string s : uset) {
        string str = s;
        if(count == 1)
            str += record;
        reverse(s.begin(), s.end());
        str += s;
        vs.push_back(str);
    }
    return vs;
}