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79. Word Search
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.
思路分析
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一刷
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二刷
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/**
* Runtime: 4 ms, faster than 89.90% of Java online submissions for Word Search.
*
* Memory Usage: 38.4 MB, less than 97.96% of Java online submissions for Word Search.
*
* Copy from: https://leetcode.com/problems/word-search/discuss/27658/Accepted-very-short-Java-solution.-No-additional-space.[Accepted very short Java solution. No additional space. - LeetCode Discuss]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-03 17:31
*/
public boolean exist(char[][] board, String word) {
if (Objects.isNull(board) || board.length == 0) {
return false;
}
if (Objects.isNull(word) || word.length() == 0) {
return false;
}
char[] chars = word.toCharArray();
for (int y = 0; y < board.length; y++) {
for (int x = 0; x < board[y].length; x++) {
if (exist(board, y, x, chars, 0)) {
return true;
}
}
}
return false;
}
private boolean exist(char[][] board, int y, int x, char[] word, int i) {
if (i == word.length) {
return true;
}
if (y < 0 || x < 0 || y == board.length || x == board[y].length) {
return false;
}
if (board[y][x] != word[i]) {
return false;
}
board[y][x] = Character.MAX_VALUE;
boolean existable = exist(board, y - 1, x, word, i + 1)
|| exist(board, y, x + 1, word, i + 1)
|| exist(board, y + 1, x, word, i + 1)
|| exist(board, y, x - 1, word, i + 1);
board[y][x] = word[i];
return existable;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-21 21:39:49
*/
public boolean exist(char[][] board, String word) {
for (int r = 0; r < board.length; r++) {
for (int c = 0; c < board[r].length; c++) {
if (board[r][c] == word.charAt(0)) {
if (backtrack(board, r, c, word, 0)) {
return true;
}
}
}
}
return false;
}
private boolean backtrack(char[][] board, int row, int column,
String word, int idx) {
if (idx >= word.length()) {
return true;
}
if (row < 0 || row >= board.length
|| column < 0 || column >= board[row].length
|| board[row][column] == ' '
|| board[row][column] != word.charAt(idx)) {
return false;
}
board[row][column] = ' ';
boolean result = backtrack(board, row - 1, column, word, idx + 1)
|| backtrack(board, row + 1, column, word, idx + 1)
|| backtrack(board, row, column - 1, word, idx + 1)
|| backtrack(board, row, column + 1, word, idx + 1);
board[row][column] = word.charAt(idx);
return result;
}