补充:
字符串的操作
str.insert() 发生重载
str.insert(str.begin(), ‘x’) 只能是插入char类型 insert(const const_iterator _Where, const _Elem _Ch)
str.insert(0, “x”) 只能是 string类型 insert(const size_type _Off, In_z const _Elem* const _Ptr)

思路:
类似字符串的切割, 传入的区间为左闭右闭 [left, right]

//自行实现, 比较复杂
class Solution {
public:
    vector<vector<string>> res;
    vector<string> res_tem;
    bool judge(string& str) {
        if (!(str.length() <= 3)) return false;
        if (str.length() > 1 && str[0] == '0') return false;
        int value = stoi(str);
        return (value >= 0 && value <= 255)? true : false;        
    }
    void myoperator(string& s, int index, int count) {
        if (res_tem.size() == 4) {
            res.push_back(res_tem);
            return;
        }
        for (int i = index; i < s.length(); i++) {
            if (count == 4) i = s.length() - 1;
            
            string str(s, index, i - index + 1);
            if (judge(str)) {
                res_tem.push_back(str);
            }   else {
                continue;
            }
            myoperator(s, i + 1, count + 1);
            res_tem.pop_back();
        }
    }
    vector<string> restoreIpAddresses(string s) {
        myoperator(s, 0, 1);
        vector<string> result;
        for (auto it = res.begin(); it != res.end(); it++) {
            string tem = "";
            for (auto i = (*it).begin(); i != (*it).end(); i++) {
                if (tem != "") tem += ".";
                tem += *i;
            }
            result.push_back(tem);
        }
        return result;
    }
};

class Solution {
public:
    vector<vector<int>> res;
    vector<int> res_tem;
    void myoperator(vector<int>& nums, int index) {
        res.push_back(res_tem);
        if (index >= nums.size()) return;

        for (int i = index; i < nums.size(); i++) {
            res_tem.push_back(nums[i]);
            myoperator(nums, i + 1);
            res_tem.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        myoperator(nums, 0);
        return res;
    }
};

思路:
同样需要进行树层去重

class Solution {
public:
    vector<vector<int>> res;
    vector<int> res_tem;
    vector<bool> used;
    void myoperator(vector<int>& nums, int index) {        
        res.push_back(res_tem);//收集元素
        if (index >= nums.size()) {
            return;
        }
        for (int i = index; i < nums.size(); i++) {
            if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == 0) {//去重操作
                continue;//此处不是break和return, 需要继续遍历
            }
            used[i] = 1;
            res_tem.push_back(nums[i]);
            myoperator(nums, i + 1);
            used[i] = 0;
            res_tem.pop_back();
        }
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        used = vector<bool>(nums.size(), 0);
        sort(nums.begin(), nums.end());
        myoperator(nums, 0);
        return res;
    }
};