return false; View in Article.
public boolean dfs(char[][] board, String word, int i, int j, int k){
}
} The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. Link: 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 are those horizontally or vertically neighboring. Subscribe. } for(int i=0; i
return true; board[i][j]=t;
board[i][j]='#';
Medium. Thanks for using LeetCode! }Similarly, below is another way of writing this algorithm.public boolean exist(char[][] board, String word) { } return false; } Subscribe to unlock. }
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for(int m=0; m<4; m++){ Approach 1: Recursion.
} } int n = board[0].length; char temp = board[i][j];
return true; } public boolean dfs(char[][] board, String word, int i, int j, int k){
if(dfs(board,word,pi,pj,k+1)){ LeetCode – Word Search (Java) Given a 2D board and a word, find if the word exists in the grid. }it means that you’ve reached the end of the word, and therefore you have found a path through the grid.y65y54yIs there any advantage of using DFS here over BFS other than DFS is easier to implement with backtracking?Why is the following check being done?Sorry got itABCCEDFB?–> should not be foundpublic boolean dfs(char[][] board, String word, int i, int j, int k){ if(i<0 || j=m || j>=n){ if(board[i][j] == word.charAt(k)){ return false;
if(dfs(board, word, i, j, 0)){ if(k==word.length()-1){ return false; }else if(dfs(board, word, i-1, j, k+1)
} result = true; The same letter cell may not be used more than once. Word Search. for(int j=0; j
} char t = board[i][j]; To view this solution you must subscribe to premium. return true;
}
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return true;
Intuition.
Hard. int m = board.length;
} The same letter cell may not be used more than once.For example, ||dfs(board, word, i, j+1, k+1)){ board[i][j]='#'; int[] di={-1,0,1,0};
Program Creek int pi=i+di[m]; int pj=j+dj[m]; 3335 164 Add to List Share.
board[i][j]=temp; return false; The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. for(int j=0; j
||dfs(board, word, i+1, j, k+1) if(board[i][j]!=word.charAt(k)){
Word Search II.
for(int i=0; i if(k>=word.length()-1){ Given a 2D board and a word, find if the word exists in the grid.
The same letter cell may not be used more than once.For example, given board =word = "ABCCED", -> returns true,This problem can be solve by using a typical DFS algorithm.public boolean exist(char[][] board, String word) {