package cn.edu.xidian.sselab.array;
/**
* title: Word Search
* content:
*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 =
*
* [
* ['A','B','C','E'],
* ['S','F','C','S'],
* ['A','D','E','E']
* ]
* word = "ABCCED", -> returns true,
* word = "SEE", -> returns true,
* word = "ABCB", -> returns false.
*/
public class WordSearch {
//这个题就像是玩贪吃蛇的游戏，抓住重点，关键点就是如何构造递归函数，以及什么时候返回值
//这个题目还有一个难点是，每个点不能重复使用，也就是说每个判断过正确的点，判断下一个节点的时候，这个值不能用，可以随便赋个值
//只要保证他的判断返回是false就可以
public boolean exist(char[][] board, String word) {
int totalLen = board.length;
int sigleLen = board[0].length;
for(int i=0;i<totalLen;i++){
for(int j=0;j<sigleLen;j++){
if(exists(board,i,j,word,0)){
return true;
}
}
}
return false;
}
private boolean exists(char[][]board,int i,int j,String word,int temp){
if(temp == word.length())
return true;
if(i<0 || j<0 || i == board.length|| j == board[0].length || board[i][j] != word.charAt(temp))
return false;
board[i][j] = '*';
boolean result = exists(board,i-1,j,word,temp+1)||
exists(board,i+1,j,word,temp+1)||
exists(board,i,j-1,word,temp+1)||
exists(board,i,j+1,word,temp+1);
board[i][j] = word.charAt(temp);
return result;
}
}