"Leetcode" Word Search (middle)

Starting today, the retrospective hardening phase.

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells is those horizontally or V Ertically neighboring. The same letter cell is used more than once.

For example,
Given board =

[  ["ABCE"],  ["SFCs"],  ["Adee"]]

Word = "ABCCED" , returns true ,
Word = "SEE" , returns true ,
Word = "ABCB" , returns false .

Ideas:

Because we already know it's backtracking, so just follow the steps.

AC, but the time is quite long 520ms

classSolution { Public:    BOOLexist (vector<vector<Char> > &board,stringword) {Vector<vector<int>>S (Word.length ()); Vector<int>X (Word.length ()); introw =board.size (); if(Row = =0)            returnWord.length () = =0; intCol = board[0].size (); //find the position of the first letter         for(inti =0; i < row; i++)        {             for(intj =0; J < Col; J + +)            {                if(Board[i][j] = = word[0]) s[0].push_back (i * col +j); }        }        intK =0; //Backtracking Solution         while(k >=0)        {             while(!S[k].empty ()) {                intLOC =S[k].back (); X[K]=Loc; intr = loc/Col; intc = loc%Col;                S[k].pop_back (); if(K < Word.length ()-1) {k= k +1; if(R-1>=0&& Board[r-1][C] = = Word[k])//above the                    {                        intLoctmp = (R-1) * col +C; if(!Isalreadyexist (X, K, loctmp))                    S[k].push_back (LOCTMP); }                    if(R +1< row && Board[r +1][C] = = Word[k])//the following                    {                        intLoctmp = (R +1) * col +C; if(!Isalreadyexist (X, K, loctmp))                    S[k].push_back (LOCTMP); }                    ifC1>=0&& Board[r][c-1] = = Word[k])//on the left.                    {                        intLoctmp = R * Col + C-1; if(!Isalreadyexist (X, K, loctmp))                    S[k].push_back (LOCTMP); }                    if(C +1< col && Board[r][c +1] = = Word[k])//on the right.                    {                        intLoctmp = R * Col + C +1; if(!Isalreadyexist (X, K, loctmp))                    S[k].push_back (LOCTMP); }                }                Else                {                    return true; }} k= K-1; }        return false; }    BOOLIsalreadyexist (vector<int> X,intKintLoc) {         for(inti =0; I < K; i++)        {            if(X[i] = =Loc)return true; }        return false; }};

The Great God 100ms Code, with recursion, in judging whether it has been used to directly find the table, will be much faster. Not find the first letter position, but unified view, the amount of code will be less

classSolution {Private: Vector<vector<Char> > *Board; string*Word; BOOL**used;Private:    BOOLIsinboard (intIintj) {if(I <0)return false; if(I >= board->size ())return false; if(J <0)return false; if(J >= (*board) [I].size ())return false; return true; }    BOOLDFS (intSiintSjintN) {if(n = = Word->size ())return true; if(Isinboard (SI, sj)) {if(!USED[SI][SJ] && (*board) [SI][SJ] = = (*word) [n]) {USED[SI][SJ]=true; BOOLRET =false; if(DFS (si+1, SJ, n+1)) RET=true; Else if(DFS (si-1, SJ, n+1)) RET=true; Else if(DFS (SI, sj+1, n+1)) RET=true; Else if(DFS (SI, sj-1, n+1)) RET=true; USED[SI][SJ]=false; returnret; }        }        return false; } Public:    BOOLexist (vector<vector<Char> > &board,stringword) {        if(board.size () = =0)return false;  This->board = &Board;  This->word = &Word; Used=New BOOL*[Board.size ()];  for(inti =0; I < board.size (); i + +) {Used[i]=New BOOL[Board[i].size ()];  for(intj =0; J < Board[i].size (); J + +) Used[i][j]=false; }         for(inti =0; I < board.size (); i + +)             for(intj =0; J < Board[i].size (); J + +)                if(DFS (I, J,0))return true; return false; }};

"Leetcode" Word Search (middle)