代码随想录day25

491.非递减子序列

//需注意此题和之前子集题的区别

    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, int start){
        if(path.size() > 1){
            result.push_back(path);
        }
        unordered_set<int> uset;
        for(int i = start; i < nums.size(); i++){
            if((!path.empty() && nums[i] < path.back()) || 
                uset.find(nums[i])!=uset.end()){
                continue;
            }
            
            path.push_back(nums[i]);
            uset.insert(nums[i]);
            backtracking(nums, i + 1);
            path.pop_back();
        }
    }
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        result.clear();
        path.clear();
        
        backtracking(nums, 0);
        return result;
    }

46.全排列

//了解排列和组合的不同

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

47.全排列II

//理解同一树层用过和同一树枝用过的不同之处

    vector<vector<int>> result;
    vector<int> path;
    
    void backtracking(vector<int>& nums, vector<bool>& used){
        if(path.size() == nums.size()){
            result.push_back(path);
            return;
        }
        for(int i = 0; i < nums.size(); i++){
            if(i>0 && (nums[i]==nums[i-1]) && (used[i-1]==false)){
                continue;
            }
            if(used[i]){
                continue;
            }
            path.push_back(nums[i]);
            used[i] = true;
            
            backtracking(nums, used);
            path.pop_back();
            used[i] = false;
        }
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        result.clear();
        path.clear();
        vector<bool> used(nums.size(), false);
        sort(nums.begin(), nums.end());
        backtracking(nums, used);
        return result;
    }