DFS template and summary

最近一直在学习Deep Frist Search，也在leetcode上解了不少的题目。从最开始的懵懂，到现在基本上遇到一个问题有了思路。我现在还清晰的接的今年2月份我刚开始刷提的时候做subsets时那个吃力的劲，脑子就是转不过来到底该如何的递归。

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;
}

对于有返回值的DFS来说。

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;
}

总结：