标签：
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.
解题思路：
开一个数组用于保存数据是否被访问过，使用DFS即可，JAVA实现如下：
static public boolean exist(char[][] board, String word) {
if (board.length == 0 || board[0].length == 0 || word.length() == 0)
return false;
boolean[][] isUsed=new boolean[board.length][board[0].length];
for (int i = 0; i < board.length; i++)
for (int j = 0; j < board[0].length; j++)
if (board[i][j] == word.charAt(0)){
isUsed[i][j]=true;
if(dfs(board,word,i,j,0,isUsed))
return true;
isUsed[i][j]=false;
}
return false;
}
public static boolean dfs(char[][] board,String word,int i,int j,int depth,boolean[][] isUsed){
if(depth==word.length()-1)
return true;
if(i>=1&&board[i-1][j]==word.charAt(depth+1)&&!isUsed[i-1][j]){
isUsed[i-1][j]=true;
if(dfs(board,word,i-1,j,depth+1,isUsed))
return true;
isUsed[i-1][j]=false;
}
if(i<=board.length-2&&board[i+1][j]==word.charAt(depth+1)&&!isUsed[i+1][j]){
isUsed[i+1][j]=true;
if(dfs(board,word,i+1,j,depth+1,isUsed))
return true;
isUsed[i+1][j]=false;
}
if(j>=1&&board[i][j-1]==word.charAt(depth+1)&&!isUsed[i][j-1]&&!isUsed[i][j-1]){
isUsed[i][j-1]=true;
if(dfs(board,word,i,j-1,depth+1,isUsed))
return true;
isUsed[i][j-1]=false;
}
if(j<=board[0].length-2&&board[i][j+1]==word.charAt(depth+1)&&!isUsed[i][j+1]){
isUsed[i][j+1]=true;
if(dfs(board,word,i,j+1,depth+1,isUsed))
return true;
isUsed[i][j+1]=false;
}
return false;
}
Java for LeetCode 079 Word Search
标签：
原文地址：http://www.cnblogs.com/tonyluis/p/4513340.html
