DFS template and summary
阿新 • • 發佈:2017-05-11
normal evel one turned set return sea 學習 ++
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