Word Search Java Solutions

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 =

[  [' A ', ' B ', ' C ', ' e '],  [' s ', ' F ', ' C ', ' s '],  [' A ', ' D ', ' e ', ' e ']]

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

Subscribe to see which companies asked

1  Public classSolution {2      Public BooleanExistChar[] board, String word) {3         if(Word = =NULL|| Word.length () = = 0)return true;4         Boolean[] Visit =New Boolean[Board.length] [Board[0].length];5         Char[] Words =Word.tochararray ();6          for(inti = 0; i < board.length; i++){7              for(intj = 0; J < Board[0].length; J + +){8                 if(BFS (BOARD,VISIT,WORDS,0,I,J))return true;9             }Ten         } One         return false; A     } -      -      Public BooleanBFS (Char[] board,Boolean[] Visit,Char[] Word,intLenintRowintCol) { the         if(len = =word.length) { -             return true; -         } -         if(Row < 0 | | row = = Board.length | | Col < 0 | | col = = board[row].length)return false; +         if(Board[row][col]! = Word[len])return false; -         if(!Visit[row][col]) { +Visit[row][col] =true; A         BooleanTMP = BFS (board,visit,word,len+1,row,col+1) | | atBFS (BOARD,VISIT,WORD,LEN+1,ROW,COL-1) | | BFS (board,visit,word,len+1,row+1,col) | | -BFS (board,visit,word,len+1,row-1, col); -Visit[row][col] =false; -         returntmp; -}Else return false; -     } in}

BFS method code logic and so on two brush time again modified learning under.

Word Search Java Solutions