搜词，一个字母只能用一遍。
比较简单，无非是看4个方向……
public boolean exist(char[][] board, String word)
{
if (board == null) return false;
int row = board.length;
int col = board[0].length;
boolean[][] visitBoard = new boolean[row][col];
for(int m = 0; m < row;m++)
{
for(int n = 0; n < col;n++)
{
visitBoard[m][n] = false;
}
}
for(int m = 0; m < row; m++)
for(int n = 0; n < col; n++)
{
if(word.charAt(0) == board[m][n])
{
visitBoard[m][n] = true;
if(helper(board,1,word,m,n,visitBoard)) return true;
visitBoard[m][n] = false;
}
}
return false;
}
public boolean helper(char[][] board, int at, String word,int m, int n,boolean[][] visitBoard)
{
if(at == word.length()) return true;
char tempChar = word.charAt(at);
boolean res = false;
//up
if( n + 1 < board[0].length)
{
if(tempChar == board[m][n+1] && !visitBoard[m][n+1])
{
visitBoard[m][n+1] = true;
res = res || helper(board,at+1,word,m,n+1,visitBoard);
visitBoard[m][n+1] = false;
}
}
//down
if( n - 1 >= 0)
{
if(tempChar == board[m][n-1] && !visitBoard[m][n-1])
{
visitBoard[m][n-1] = true;
res = res || helper(board,at+1,word,m,n-1,visitBoard);
visitBoard[m][n-1] = false;
}
}
//left
if( m-1 >= 0)
{
if(tempChar == board[m-1][n] && !visitBoard[m-1][n])
{
visitBoard[m-1][n] = true;
res = res || helper(board,at+1,word,m-1,n,visitBoard);
visitBoard[m-1][n] = false;
}
}
//right
if( m+1 < board.length)
{
if(tempChar == board[m+1][n] && !visitBoard[m+1][n])
{
visitBoard[m+1][n] = true;
res = res || helper(board,at+1,word,m+1,n,visitBoard);
visitBoard[m+1][n] = false;
}
}
return res;
}
写的很乱 改了几次 感觉完全可以写的更好
再说吧。。
二刷。
幸亏二刷昨晚比一刷的描述要好，否则没有进步真让我觉得自己是真蠢。True Dumb..
BackTrack吧，从每个点开始都试一次，boolean[][]标记使用过的元素，go deeper的条件是 找到 + deeper direction unvisited and in range
public class Solution
{
public boolean exist(char[][] board, String word)
{
if(board.length == 0) return false;
boolean[][] check = new boolean[board.length][board[0].length];
for(int i = 0; i < board.length;i++)
{
for(int j = 0; j < board[0].length;j++)
{
check[i][j] = true;
if(helper(i,j,board,word,check,0)) return true;
check[i][j] = false;
}
}
return false;
}
public boolean helper(int x, int y,char[][] board, String word, boolean[][] check,int m)
{
if(m == word.length()) return true;
else
{
if(board[x][y] == word.charAt(m))
{
if(m == word.length()-1) return true;
if(x > 0 && !check[x-1][y])
{
check[x-1][y] = true;
if(helper(x-1,y,board,word,check,m+1)) return true;
check[x-1][y] = false;
}
if(y > 0 && !check[x][y-1])
{
check[x][y-1] = true;
if(helper(x,y-1,board,word,check,m+1)) return true;
check[x][y-1] = false;
}
if(x < board.length-1 && !check[x+1][y])
{
check[x+1][y] = true;
if(helper(x+1,y,board,word,check,m+1)) return true;
check[x+1][y] = false;
}
if(y < board[0].length-1 && !check[x][y+1])
{
check[x][y+1] = true;
if(helper(x,y+1,board,word,check,m+1)) return true;
check[x][y+1] = false;
}
return false;
}
else return false;
}
}
}
BackTrack的题大多数比较基础，属于pure programming skill,基本和数学无关呀。。
三刷。
2D板子上进行DFS的题很多。。
三刷学着把DFS写简单点，主要是要在DFS的开头进行条件判断。
然后如果棋盘不是read only可以在棋盘上标记是否访问过，换成别的CHAR之类的再换回来。
然后如果多一个变量记录比较到WORD中哪一个字母，可以不用反复做substring，substring在java里很不友好。。
Time: O((mn)4^k) 感觉是这个，每次4个方向，depth是单词长度K，然后每个位置都要来一次。。
Space: O(mn) 不是read only的话可以O(1)？
public class Solution {
public boolean exist(char[][] board, String word) {
if (board.length == 0 || word.length() == 0) return false;
boolean[][] visited = 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(dfs(board, visited, i, j, word)) {
return true;
}
visited[i][j] = false;
}
}
return false;
}
public boolean dfs(char[][] board, boolean[][] visited, int m, int n, String word) {
if (word.length() == 0) return true;
if (m < 0 || m == board.length || n < 0 || n == board[0].length || visited[m][n] || board[m][n] != word.charAt(0)) return false;
visited[m][n]= true;
if (dfs(board, visited, m+1, n, word.substring(1)) || dfs(board, visited, m, n+1, word.substring(1)) ||
dfs(board, visited, m-1, n, word.substring(1)) || dfs(board, visited,m, n-1, word.substring(1))) {
return true;
} else {
visited[m][n] = false;
return false;
}
}
}