LeetCode | Word Search

题目

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.

For example,

Given board =

[
["ABCE"],
["SFCS"],
["ADEE"]
]

word = 

"ABCCED"

,
-> returns 

true

,
word = 

"SEE"

,
-> returns 

true

,
word = 

"ABCB"

,
-> returns 

false

.
分析

矩阵搜索的一般就考虑BFS或DFS了，BFS的方法写起来代码较长，是要纪录每一层符合条件的元素的位置信息。这里给出的是DFS的，其实就是非递归不好写，写个递归的了。

代码

public class WordSearch {
private char[] wordCharArray;
private int L;
private char[][] board;
private int M;
private int N;
private boolean[][] mark;

public boolean exist(char[][] board, String word) {
if (word == null || word.length() == 0 || board == null
|| board.length == 0) {
return false;
}
wordCharArray = word.toCharArray();
L = wordCharArray.length;
this.board = board;
M = board.length;
N = board[0].length;
mark = new boolean[M]
;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (dfs(i, j, 0)) {
return true;
}
}
}
return false;
}

private boolean dfs(int i, int j, int k) {
if (i < 0 || j < 0 || i >= M || j >= N || mark[i][j]) {
return false;
}
if (board[i][j] != wordCharArray[k]) {
return false;
}
if (k == L - 1) {
return true;
}
mark[i][j] = true;
boolean result = dfs(i - 1, j, k + 1) || dfs(i + 1, j, k + 1)
|| dfs(i, j - 1, k + 1) || dfs(i, j + 1, k + 1);
mark[i][j] = false;
return result;
}
}