全排列问题：
path
递归终止条件：path中是否已存储所有元素；
for循环处理节点集合：used=0未被使用的元素

class Solution {
public:vector<int> path;vector<vector<int>> res;void backtracking(vector<int>& nums,vector<bool>& used) {if(path.size() == nums.size()){res.push_back(path);return;}for(int i=0;i<nums.size();i++) {if(used[i]==true)continue;path.push_back(nums[i]);used[i]=true;backtracking(nums,used);used[i]=false;path.pop_back();}}vector<vector<int>> permute(vector<int>& nums) {vector<bool> used(nums.size(),false);backtracking(nums,used);return res;}
};

子集问题：
每个集合都需要存储到res中，
递归终止条件：可以省略，startIndex下标到达数组末尾 与for循环的条件判断一致
for循环处理的元素集合，startIndex-数组最后一个元素

class Solution {
public:vector<vector<int>> res;vector<int> path;void backtracking(vector<int>& nums,int startIndex) {res.push_back(path);for(int i=startIndex;i<nums.size();i++) {path.push_back(nums[i]);backtracking(nums,i+1);path.pop_back();}}vector<vector<int>> subsets(vector<int>& nums) {backtracking(nums,0);return res;}
};

这个题根据digits中给出的数字字符，从而锁定对应的字符串，对字符串中的字母进行组合，前面复习过两遍，对这个题的印象还不是很清晰，故再写一次题解：

class Solution {const string strSet[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
public:string s;vector<string> res;void backtracking(const string& digits,int index) {if(index == digits.size()) {res.push_back(s);return;}int digit = digits[index]-'0';string str = strSet[digit];for(int i=0;i<str.size();i++) {s.push_back(str[i]);backtracking(digits,index+1);s.pop_back();}}vector<string> letterCombinations(string digits) {if(digits.size()==0)return res;backtracking(digits,0);return res;}
};

元素可以多次使用，下一次元素集合的开始位置可以还是startIndex

class Solution {
public:vector<vector<int>> res;vector<int> path;void backtracking(vector<int>& candidates,int target,int sum,int startIndex) {if(sum > target)return;if(sum == target) {res.push_back(path);return;}for(int i=startIndex;i<candidates.size();i++) {path.push_back(candidates[i]);sum+=candidates[i];backtracking(candidates,target,sum,i);sum-=candidates[i];path.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {int sum=0;backtracking(candidates,target,sum,0);return res;}
};

/*
// Definition for a Node.
class Node {
public:int val;Node* next;Node* random;Node(int _val) {val = _val;next = NULL;random = NULL;}
};
*/class Solution {
public:unordered_map<Node*,Node*> cachedNode;Node* copyRandomList(Node* head) {if(head == nullptr)return nullptr;//如果当前节点没有被拷贝if(!cachedNode.count(head)) {//新定义节点headNew保存val,Node* headNew = new Node(head->val);cachedNode[head] = headNew;//head->next 和head->random要先存在了才能指过去//利用递归完成headNew的后继节点next和随机指向节点random的拷贝headNew->next = copyRandomList(head->next);headNew->random = copyRandomList(head->random);}return cachedNode[head];}
};