Leetcode 46. Permutations 排列 回溯法典型应用

• 兄弟题 LQ 47 带重的回溯
• 值得学习的点 回溯法框架
• vector 删除最后一个元素 pop_back(), 还有erase()和substring
  • erase 可以删除指定位置元素，或者一定范围的元素

    •   C++11 erase()
       iterator erase (const_iterator position);
       iterator erase (const_iterator first, const_iterator last);
      

    • s.substring(起点，保留的字符数)

code

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<int> track;
        backtrack(nums, track);
        return res;
    }
private:
    vector<vector<int>> res;
    void backtrack(vector<int>& nums, vector<int>& track){
        if(track.size()==nums.size()){
            res.push_back(track);
            return ;
        }
        for(auto& num: nums){
            bool flag=1;
            for(auto& tmp: track){
                if(tmp==num) 
                    flag=0;
            }
            if(flag==0) continue;
            track.push_back(num);
            backtrack(nums, track);
            track.pop_back();
        }
        return ;
    }
};

code 2

• 这里可以用一个used 布尔变量数组，来记录每个数是不是被用过了

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        paths.clear();
        path.clear();

        vector<int> used(nums.size(),false);

        helper(nums,used);

        return paths;
    }
private:
    void helper(vector<int>& nums,vector<int>& used) {
        if(path.size() == nums.size()) {
            paths.push_back(path);
            return ;
        }

        for(int i = 0 ; i < nums.size() ; ++i) {
            if(used[i]) continue;

            used[i] = true;
            path.push_back(nums[i]);
            
            helper(nums,used);

            path.pop_back();
            used[i] = false;
        }
    }
private:
    vector<vector<int>> paths;
    vector<int> path;
};

code3

• 避免动态修改数组

class Solution {
public:
   vector<vector<int>> res;
   int n;
   vector<int> path;
   vector<int> used;
   vector<vector<int>> permute(vector<int>& nums) {
       n = nums.size();
       path.resize(n, -1);
       used.resize(n, false);
       backtrack(nums, 0);        
       return res;
   }

   void backtrack(vector<int> &nums, int t){

       if(t == n) {
           res.push_back(path);
           return ;
       }
       for(int i = 0; i < n; i ++){
           if(used[i] == false){
               used[i] = true;
               path[t] = nums[i];
               backtrack(nums, t + 1);
               used[i] = false;
           }
       }
   }
};