第七章 回溯演算法part01

晴夜空發表於2024-07-13
演算法

第七章 回溯演算法part01

77. 組合 216.組合總和III 17.電話號碼的字母組合

77. 組合

題目連結 :

77. 組合 - 力扣（LeetCode）

Code ( 用 遞迴 模擬 迴圈 ) : 

class Solution {
public:
    vector<vector<int>> combine(int n, int k) {
​
        vector<int> vec_Temp_Cache_For_RecordNum ;
​
        vector<vector<int>> vec_2_Dimention_For_Receive_And_Return ;
​
        work_Recursino_For_forLoop( 1 , 1 , n , k , vec_Temp_Cache_For_RecordNum , vec_2_Dimention_For_Receive_And_Return  )  ;
​
​
        return vec_2_Dimention_For_Receive_And_Return ;
​
​
​
    }
​
​
​
​
    void work_Recursino_For_forLoop( int i_Work_In , int i_Layer , int n , int k , vector<int> & vec_Temp_Cache_For_RecordNum , vector<vector<int>> & vec_2_Dimention_For_Receive_And_Return )
    {
​
        if(i_Layer != k )       // 邏輯 性     /   安全 性       /       效率    
        {
            for( int i = i_Work_In; i <= (( n - k ) + i_Layer ) ; i++ )
            {
                //vec_Temp_Cache_For_RecordNum[ (i_Layer - 1 ) ] = i ; 
​
                vec_Temp_Cache_For_RecordNum.push_back(i) ;
​
​
                work_Recursino_For_forLoop( (i + 1 ) , ( i_Layer + 1  ) , n , k , vec_Temp_Cache_For_RecordNum , vec_2_Dimention_For_Receive_And_Return  )  ;
​
                vec_Temp_Cache_For_RecordNum.pop_back() ;
​
​
            }
            
​
​
​
        }
        else
        {
            for( int i = i_Work_In; i <= (( n - k ) + i_Layer ) ; i++ )
            {
​
                //vec_Temp_Cache_For_RecordNum[ (i_Layer - 1 ) ] = i ;
​
                vec_Temp_Cache_For_RecordNum.push_back(i) ;
​
                vec_2_Dimention_For_Receive_And_Return.push_back(vec_Temp_Cache_For_RecordNum ) ;
​
                //work_Recursino_For_forLoop( (i + 1 ) , ( i_Layer + 1  ) , n , k  )
​
                vec_Temp_Cache_For_RecordNum.pop_back() ;
​
​
            }
​
​
​
        }
​
        
​
​
    }
};

216.組合總和III

題目連結 :

216. 組合總和 III - 力扣（LeetCode）

Code ( 遞迴 遍歷 , 回溯 , 不斷 地 更新 Cache 向量 中 的 值 , Cache Sum ) : 

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
​
        int target = n ;
​
​
        vector<int> vec_Temp_Cache_For_NumSequence ;
​
        vector<vector<int>> vec_2_Dimention_For_Receive_And_Return ;
​
        int num_Cache_Sum = 0 ;
​
​
​
        // 嘗試 遞迴  遍歷 
​
​
​
        fuction_Work_Recursion_For_forLoop(1 , 1 , k , target , vec_Temp_Cache_For_NumSequence , vec_2_Dimention_For_Receive_And_Return , num_Cache_Sum   ) ;  
​
​
        return vec_2_Dimention_For_Receive_And_Return ; 
​
​
​
​
​
    }
​
​
​
​
    void fuction_Work_Recursion_For_forLoop( int i_Work_In , int i_Layer , int k , int target , vector<int> & vec_Temp_Cache_For_NumSequence , vector<vector<int>> & vec_2_Dimention_For_Receive_And_Return , int & num_Cache_Sum    )
    {                                                                                                                                                                                                   // “ 省 記憶體  ”  , Question :  會  省  暫存器  嗎 ?  // 程式碼 會不會  自動 地   進行   相關   最佳化   
​
        if( i_Layer != k )
        {
            for( int i = i_Work_In ; i <= (( 9 - k ) + i_Layer ) ; i++  )
            {
                
                num_Cache_Sum += i ; 
​
                vec_Temp_Cache_For_NumSequence.push_back(i) ;
​
                //
​
                fuction_Work_Recursion_For_forLoop( ( i + 1 ) , ( i_Layer + 1 ) , k , target , vec_Temp_Cache_For_NumSequence , vec_2_Dimention_For_Receive_And_Return , num_Cache_Sum     )     ;
​
                
                num_Cache_Sum -= i ;
​
                vec_Temp_Cache_For_NumSequence.pop_back() ;
​
​
​
            }
​
​
​
        }
        else
        {
            for( int i = i_Work_In ; i <= (( 9 - k ) + i_Layer ) ; i++  )
            {
​
                num_Cache_Sum += i ; 
​
                vec_Temp_Cache_For_NumSequence.push_back(i) ;
​
                //
​
                if(num_Cache_Sum == target )
                {
​
                    vec_2_Dimention_For_Receive_And_Return.push_back(vec_Temp_Cache_For_NumSequence ) ;
​
​
                }
                
                
​
                num_Cache_Sum -= i ;
                
                vec_Temp_Cache_For_NumSequence.pop_back() ;
​
​
​
​
​
​
            }
​
​
​
        }
​
        
​
​
​
    }
​
​
};

17.電話號碼的字母組合

題目地址 :

17. 電話號碼的字母組合 - 力扣（LeetCode）

Code ( 版本 1 ( Cache 字串 引用 + 維護 ) ) : 

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        
        vector<vector<string>> vec_Resource_Num_Map_Letter = { { "a" , "b" , "c" }, { "d" , "e" , "f" } , {"g","h","i"} , {"j","k","l"} , {"m","n","o"} , {"p" , "q" , "r" , "s" } , { "t" , "u" , "v"} , { "w" , "x" , "y" , "z"  } } ;
​
        //vec_Resource_Num_Map_Letter
​
        string str_Cache = "" ;
​
        vector<string> vec_Str_For_Receive_And_Return ;
​
​
​
        if( digits == "")
        {
​
            return vec_Str_For_Receive_And_Return ;
​
​
        }
​
​
        fuction_Work_Recursion_For_forLoop( 0 , str_Cache , vec_Str_For_Receive_And_Return , vec_Resource_Num_Map_Letter , digits  )  ;      
​
​
​
        return vec_Str_For_Receive_And_Return ;
​
    }
​
​
    void fuction_Work_Recursion_For_forLoop(int i_Layer , string & str_Cache , vector<string> & vec_Str_For_Receive_And_Return , vector<vector<string>> & vec_Resource_Num_Map_Letter , string digits  )
    {
        char char_Cache ;
        
​
        if( digits[i_Layer + 1] != '\0' )
        {
            char_Cache = digits[i_Layer ] ;
​
            int num_Temp_Cache = int ((char_Cache - 0x30 ) - 2 ) ;
​
​
            for(int i = 0 ; i < vec_Resource_Num_Map_Letter[num_Temp_Cache].size() ; i++ )
            {
                str_Cache += vec_Resource_Num_Map_Letter[num_Temp_Cache][i] ;
​
​
                //
​
                fuction_Work_Recursion_For_forLoop( (i_Layer + 1  ) , str_Cache , vec_Str_For_Receive_And_Return , vec_Resource_Num_Map_Letter , digits  )   ;    
​
​
                str_Cache = str_Cache.substr(0 , i_Layer ) ;
​
​
            }
​
​
​
        }
        else
        {
            char_Cache = digits[i_Layer ] ;
​
            int num_Temp_Cache = int ((char_Cache - 0x30 ) - 2 ) ;
​
​
            for(int i = 0 ; i < vec_Resource_Num_Map_Letter[num_Temp_Cache].size() ; i++ )
            {
                str_Cache += vec_Resource_Num_Map_Letter[num_Temp_Cache][i] ;
​
​
                //
                vec_Str_For_Receive_And_Return.push_back(str_Cache ) ;
​
​
                str_Cache = str_Cache.substr(0 , i_Layer ) ;
​
​
            }
​
​
​
​
​
        }
​
​
​
​
​
    }
​
​
​
​
};

Code ( 版本 2 ( Cache 字串 傳值) ) : 

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        
        vector<vector<string>> vec_Resource_Num_Map_Letter = { { "a" , "b" , "c" }, { "d" , "e" , "f" } , {"g","h","i"} , {"j","k","l"} , {"m","n","o"} , {"p" , "q" , "r" , "s" } , { "t" , "u" , "v"} , { "w" , "x" , "y" , "z"  } } ;
​
        //vec_Resource_Num_Map_Letter
​
        string str_Cache = "" ;
​
        vector<string> vec_Str_For_Receive_And_Return ;
​
​
​
        if( digits == "")
        {
​
            return vec_Str_For_Receive_And_Return ;
​
​
        }
​
​
        fuction_Work_Recursion_For_forLoop( 0 , str_Cache , vec_Str_For_Receive_And_Return , vec_Resource_Num_Map_Letter , digits  )  ;      
​
​
​
        return vec_Str_For_Receive_And_Return ;
​
    }
​
​
    void fuction_Work_Recursion_For_forLoop(int i_Layer , string str_Cache , vector<string> & vec_Str_For_Receive_And_Return , vector<vector<string>> & vec_Resource_Num_Map_Letter , string digits  )
    {
        char char_Cache ;
        
​
        if( digits[i_Layer + 1] != '\0' )
        {
            char_Cache = digits[i_Layer ] ;
​
            int num_Temp_Cache = int ((char_Cache - 0x30 ) - 2 ) ;
​
​
            for(int i = 0 ; i < vec_Resource_Num_Map_Letter[num_Temp_Cache].size() ; i++ )
            {
                //str_Cache += vec_Resource_Num_Map_Letter[num_Temp_Cache][i] ;
​
​
                //
​
                fuction_Work_Recursion_For_forLoop( (i_Layer + 1  ) , (str_Cache + vec_Resource_Num_Map_Letter[num_Temp_Cache][i] ) , vec_Str_For_Receive_And_Return , vec_Resource_Num_Map_Letter , digits  )   ;    
​
​
                //str_Cache = str_Cache.substr(0 , i_Layer ) ;
​
​
            }
​
​
​
        }
        else
        {
            char_Cache = digits[i_Layer ] ;
​
            int num_Temp_Cache = int ((char_Cache - 0x30 ) - 2 ) ;
​
​
            for(int i = 0 ; i < vec_Resource_Num_Map_Letter[num_Temp_Cache].size() ; i++ )
            {
                //str_Cache += vec_Resource_Num_Map_Letter[num_Temp_Cache][i] ;
​
​
                //
                vec_Str_For_Receive_And_Return.push_back(str_Cache + vec_Resource_Num_Map_Letter[num_Temp_Cache][i] ) ;
​
​
                //str_Cache = str_Cache.substr(0 , i_Layer ) ;
​
​
            }
​
​
​
​
​
        }
​
​
​
​
​
    }
​
​
​
​
};

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