class Solution {
public:vector<string> res;bool isVaild(const string &s,int start,int end){//闭区间if(start>end)return false;if(s[start]=='0'&&start!=end){return false;}int num=0;for(int i=start;i<=end;i++){if(s[i]>'9'||s[i]<'0')return false;num = num * 10 + (s[i] - '0');if(num>255)return false;}return true;}void backtracking(string &s,int startIndex,int pointNum){if(pointNum==3){//判断第四段是否符合要求if(isVaild(s,startIndex,s.size()-1))res.push_back(s);return;}for(int i=startIndex;i<s.size();i++){if(isVaild(s,startIndex,i)){s.insert(s.begin()+i+1,'.');backtracking(s,i+2,pointNum+1);s.erase(s.begin()+i+1);}else{break;//说明所有的都不合法了直接break}}}vector<string> restoreIpAddresses(string s) {backtracking(s,0,0);return res;}
};

思路：

递归遍历 pointNum=3为终止条件 isValue判断是不是符合要求闭区间得证

class Solution {
public:vector<vector<int>> res;vector<int> path;void backtracking(vector<int>& nums,int startIndex){if(startIndex==nums.size()){res.push_back(path);return;} 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;}
};

class Solution {
public:vector<vector<int>> res;vector<int> path;void backtracking(vector<int>& nums,int startIndex,vector<bool> is){if(startIndex==nums.size()){res.push_back(path);return;}res.push_back(path);for(int i=startIndex;i<nums.size();i++){//false代表是同一树层 现在属于前面的真子集一定被遍历过了 直接continueif(i>0&&nums[i]==nums[i-1]&&is[i-1]==false){continue;}path.push_back(nums[i]);is[i]=true;backtracking(nums,i+1,is);is[i]=false;path.pop_back();}}vector<vector<int>> subsetsWithDup(vector<int>& nums) {vector<bool> is(nums.size(),false);sort(nums.begin(),nums.end());backtracking(nums,0,is);return res;}
};