93. 复原 IP 地址

不断向s中插入'.'，然后回溯删除

class Solution {
private:
    vector<string> result;
    int count = 0;
    void back_tracking(string& s,int start,int count){
        if(count == 3){
            if(validate(s,start,s.size()-1))
                result.push_back(s);
        }
        for(int i = start; i < s.size();i++){
            if(validate(s, start, i)){
                s.insert(s.begin() + i + 1, '.');
                count++;
                back_tracking(s, i + 2, count);
                count--;
                s.erase(s.begin() + i + 1);
            }
        }
    }
    bool validate(const string& s, int start,int end){
        if (start > end) {
            return false;
        }
        if (s[start] == '0' && start != end) { // 0开头的数字不合法
            return false;
        }
        int num = 0;
        for (int i = start; i <= end; i++) {
            if (s[i] > '9' || s[i] < '0') { // 遇到非数字字符不合法
                return false;
            }
            num = num * 10 + (s[i] - '0');
            if (num > 255) { // 如果大于255了不合法
                return false;
            }
        }   
        return true;
    }
public:
    vector<string> restoreIpAddresses(string s) {
        back_tracking(s, 0, 0);
        return result;
    }
};

78. 子集

 1 class Solution {
 2 private:
 3     vector<vector<int>> result;
 4     vector<int> path;
 5     void back_tracking(const vector<int>& nums, int start){
 6         result.push_back(path);
 7         for(int i = start; i < nums.size(); i++){
 8             path.push_back(nums[i]);
 9             back_tracking(nums, i + 1);
10             path.pop_back();
11         }
12     }
13 public:
14     vector<vector<int>> subsets(vector<int>& nums) {
15         back_tracking(nums, 0);
16         return result;
17     }
18 };

90. 子集 II

声明一个去重的数组used，先排序，然后回溯

 1 class Solution {
 2 private:
 3     vector<vector<int>> result;
 4     vector<int> path;
 5     void back_tracking(const vector<int>& nums, int start, vector<bool>& used){
 6         result.push_back(path);
 7         for(int i = start; i < nums.size(); i++){
 8             if(i > 0 && nums[i] == nums[i-1] && used[i-1] == false) continue;
 9             path.push_back(nums[i]);
10             used[i] = true;
11             back_tracking(nums, i+1, used);
12             used[i] = false;
13             path.pop_back();
14         }
15     }
16 public:
17     vector<vector<int>> subsetsWithDup(vector<int>& nums) {
18         vector<bool> used(nums.size(),false);
19         sort(nums.begin(), nums.end());
20         back_tracking(nums, 0, used);
21         return result;
22     }
23 };