Leetcode@word search
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For this problem, it is a common depth first search problem, and use the visited boolean array to judge if the char element has been visited before.
one problem I meet if that start means the number of the word, when I finish the matching I should return true in this recursive loop not the next one. Because the program may not enter the next loop, so it can not enter the next recursive loop.
another problem, when we do the backtracking, for this problem it is to delete the turn the visited element from true to false.
public class Solution {
public boolean exist(char[][] board, String word) {int length = word.length();
if(length <= 0)
return true;
int rowNum = board.length;
if(rowNum == 0)
return false;
int columnNum = board[0].length;
if(columnNum == 0)
return false;
boolean[][] visited = new boolean[rowNum][columnNum];
for(int i = 0; i < rowNum; i++){
for(int j = 0; j < columnNum; j++){
if(helper(word, 0, board, i, j,visited))
return true;
}
}
return false;
}
public boolean helper(String s, int start, char[][]board, int row, int column,boolean[][] visited){
int length = s.length();
if(visited[row][column])
return false;
if(s.charAt(start) != board[row][column])
return false;
visited[row][column] = true;
if(start+1 >= length){
visited[row][column] = false;
return true;
}
int rowNum = board.length;
int columnNum = board[0].length;
if(row + 1< rowNum && helper(s,start + 1,board, row+1,column,visited)){
visited[row][column] = true;
return true;
}
if(row - 1>= 0 && helper(s,start + 1,board, row-1,column,visited)){
visited[row][column] = false;
return true;
}
if(column + 1< columnNum && helper(s,start + 1,board, row,column + 1,visited)){
visited[row][column] = false;
return true;
}
if(column - 1>= 0 && helper(s,start + 1,board, row,column - 1,visited)){
visited[row][column] = false;
return true;
}
visited[row][column] = false;
return false;
}
}
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