为给定字符串生成所有可能的回文

Inputs : I.1) A string of characters
Outputs: O.1) Total number of palindromes found
         O.2) Palindromes output on standard output
Note : duplicate palindromes are not allowed
             output need not be sorted
single characters are a palindrome.

我已经通过回溯解决了上述问题，但它为一个大的输入字符串提供了 TLE。因此，如果解决方案避免递归，这将是有益的。TLE 的输入是 ：- amitabhbacchan
这是我的实现：

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
int isSafe(int *a, int i) {
    if (a[i] == 0) {
        return 1;
    }
    else {
        return 0;
    }
}
int isPalin(string b) {
    if (b.length() == 0) {
        return 0;
    }
    if (b.length() == 1) {
        return 1;
    }
    string c = b;
    std::reverse(c.begin(), c.end());
    if (c == b) {
        return 1;
    }
    else {
        return 0;
    }
}
void palindrome(string a, string b, int *vec, std::set<string>&res) {
    if (isPalin(b)) {
        if (res.find(b) != res.end()) {
            return;
        }
        res.insert(b);
    }
    for (int i = 0; i < a.length(); i++) {
        if (isSafe(vec, i)) {
            b.push_back(a[i]);
            vec[i] = 1;
            palindrome(a, b, vec, res);
            b.pop_back();
            vec[i] = 0;
        }
    }
}
int main(void) {
    string a = "ababcdef";
    string c = "";
    int *vec;
    std::set<string>res;
    vec = (int*)malloc(a.length()*sizeof(int));
    for (int i = 0; i < a.length(); i++) {
        vec[i] = 0;
    }
    palindrome(a, c, vec, res);
    std::set<string>::iterator it;
    for (it = res.begin(); it != res.end(); it++) {
        cout << *it << endl;
    }
    return 0;
}