最近一直在學習Deep Frist Search，也在leetcode上解了不少的題目。從最開始的懵懂，到現在基本上遇到一個問題有了思路。我現在還清晰的接的今年2月份我剛開始刷提的時候做subsets時那個吃力的勁，腦子就是轉不過來到底該如何的遞歸。

void DFS_no_return(vector<TYPE>& solution_set, TYPE& solution, int idx){
    // DFS terminate condition
    // Normally there are two major categories
    
 
// One is for some number, already reach this depth according to the question, stop
    // get to the end of an array, stop
    if(idx == n || idx == (vector/string).size()){
        solution_set.push_back(solution);
        return;
    }
    // very seldom only when our program met some efficiency problem
    
 
// do prune to reduce the workload
    // no need do DFS if we already get the value of current element, just return
    if(hash_table[input] != hash_table.end()){
        return;
    }
    
    for(int i = 0; i < n; i++){
        if(visited[i] == 0){
            visited[i] = 1;
            DFS_no_return(solution_set, solution, idx 
 
+ 1);
            visited[i] = 0;
        }
    }
    return;
}

對於有返回值的DFS來說。

TYPE DFS_with_return(int idx){
    // DFS terminate condition
    // Normally there are two major categories
    // One is for some number, already reach this depth according to the question, stop
    // get to the end of an array, stop
    if(idx == n || idx == (vector/string).size()){
        return 1/0/"";
    }
    // very seldom only when our program met some efficiency problem
    // do prune to reduce the workload
    // no need do DFS if we already get the value of current element, just return
    if(hash_table[input] != hash_table.end()){
        return hash_table[input];
    }
    
    TYPE ans;
    for(int i = 0; i < n; i++){
        if(visited[i] == 0){
            // normally, here should to add/concatenate the value in current level with the value returned from lower level
            ans += DFS_no_return(idx + 1);
        }
    }
    // if we need prune
    // keep current value here
    hash_table[input] = ans;
    return ans;
}

總結：

DFS template and summary