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79.word-search.java
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74 lines (68 loc) · 1.86 KB
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/*
* @lc app=leetcode id=79 lang=java
*
* [79] Word Search
*
* https://leetcode.com/problems/word-search/description/
*
* algorithms
* Medium (32.77%)
* Likes: 2375
* Dislikes: 126
* Total Accepted: 347.4K
* Total Submissions: 1.1M
* Testcase Example: '[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]]\n"ABCCED"'
*
* 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. The
* same letter cell may not be used more than once.
*
* Example:
*
*
* board =
* [
* ['A','B','C','E'],
* ['S','F','C','S'],
* ['A','D','E','E']
* ]
*
* Given word = "ABCCED", return true.
* Given word = "SEE", return true.
* Given word = "ABCB", return false.
*
*
*/
// @lc code=start
class Solution {
public boolean exist(char[][] board, String word) {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
if (board[i][j] == word.charAt(0) && dfs(board, i, j, 0, word)){
return true;
}
}
}
return false;
}
// DFS approach
public boolean dfs(char board[][], int i, int j, int count, String word) {
if (count == word.length()) {
return true;
}
if (i < 0 || i >= board.length || j < 0 || j >= board[i].length || board[i][j] != word.charAt(count)){
return false;
}
char temp = board[i][j];
board[i][j] = ' ';
boolean found = dfs(board, i+1, j, count+1, word)
|| dfs(board, i-1, j, count+1, word)
|| dfs(board, i, j+1, count+1, word)
|| dfs(board, i, j-1, count+1, word);
board[i][j] = temp;
return found;
}
}
// @lc code=end