题解 | #字符串的排列#

#include <iterator>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param str string字符串 
     * @return string字符串vector
     */
    void backtrace(string str, string current, vector<string> &res, vector<bool> &used)
    {
        int n1 = str.length(), n2 = current.length();
        if(n1 == n2)
        {
            res.push_back(current);
            return;
        }

        for(int i = 0; i < n1; ++i)
        {
            if(used[i] || (i > 0 && str[i] == str[i - 1] && !used[i - 1]))
                continue;

            used[i] = true;
            current += str[i];
            backtrace(str, current, res, used);
            used[i] = false;
            current.pop_back();
        }
    }
    
    vector<string> Permutation(string str) {
        if(str == "")
            return {};

        string current = "";
        vector<string> res;
        int n = str.length();
        vector<bool> used(n, false);
        sort(str.begin(), str.end());

        backtrace(str, current, res, used);

        return res;
    }
};

跟有重复数字的数列的全排列是一个思路，不多说了