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全排列问题是回溯法的典型例子。可行解的组成形式是给定数组中的所有数的组合，因此可作为可行解的判定条件。每次需要在剩下可被选中的集合中选择一个，创建mask数组。

class Solution {public:    void dfs(vector
   
     vct, vector
    
      cur, const vector
     
      & nums, vector
      
       & used) {        if (cur.size() == nums.size()) {            vct.push_back(cur);            return;        }        for (int i = 0; i < nums.size(); ++i) {            if (used[i] == 0) {                cur.push_back(nums[i]);                used[i] = 1;                dfs(vct, cur, nums, used);                used[i] = 0;                cur.pop_back();            }        }    }    vector
       
        
         > permute(const vector
         
          & nums) { vector
          
           
            > vct; if (nums.size() <= 0) return vct; vector
            
              cur; vector
             
               used(nums.size(), false); dfs(vct, cur, nums, used); return vct; }};
             
            
           
          
         
        
       
      
     
    
   

diff：需要考虑val1 = val2的情况，需排序将相同元素聚类，参考前文处理重复的方法。

class Solution {public:    void dfs(vector
   
     vct, vector
    
      cur, const vector
     
      & nums, vector
      
       & used) {        if (cur.size() == nums.size()) {            vct.push_back(cur);            return;        }        for (int i = 0; i < nums.size(); ++i) {            if (used[i]) continue;            int pre_index = i - 1;            bool repeated = false;            while (pre_index >= 0 && nums[pre_index] == nums[i]) {                if (used[pre_index]) {                    repeated = true;                    break;                }                --pre_index;            }            if (repeated) continue;            cur.push_back(nums[i]);            used[i] = true;            dfs(vct, cur, nums, used);            used[i] = false;            cur.pop_back();        }    }    vector
       
        
         > permuteUnique(const vector
         
          & nums) { vector
          
           
            > vct; if (nums.size() <= 0) return vct; vector
            
              cur; vector
             
               used(nums.size(), false); dfs(vct, cur, nums, used); return vct; }};
             
            
           
          
         
        
       
      
     
    
   

注意：请根据实际需求补充具体内容，确保代码符合上下文，并维护功能性和编译正确性。