79-WordSearch

【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.

【analyze】

1.类似迷宫问题，采用DFS

2.需要一个辅助二维数组，记录遍历过的位置

3.需要注意：当某个位置置为true后，没有成功后需要重置为false

【算法】public class Solution {

public boolean exist(char[][] board, String word) {
        boolean[][] visited=new boolean[board.length][board[0].length];  //用于判断一个点是否遍历过了
        for(int i=0;i<board.length;i++) {
            for(int j=0;j<board[0].length;j++) {
                if(board[i][j]==word.charAt(0)) {
                    visited[i][j]=true;
                    if(word.length()==1||search(board,i,j,word.substring(1),visited))
                        return true;
                    visited[i][j]=false;
                }
            }
        }
        return false;
    }

    public boolean search(char[][] board,int i,int j,String word,boolean[][] visited) {
        if(word.length()==0)
            return true;
        int[][] direction={{0,1},{0,-1},{1,0},{-1,0}};    //四个方向
        for(int k=0;k<direction.length;k++) {
            int ii=i+direction[k][0];
            int jj=j+direction[k][1];
            if(ii>=0&&ii<board.length&&jj>=0&&jj<board[0].length&&board[ii][jj]==word.charAt(0)&&visited[ii][jj]==false) {
                visited[ii][jj]=true;
                if(word.length()==1||search(board,ii,jj,word.substring(1),visited))
                    return true;
                visited[ii][jj]=false;
            }
        }
        return false;
    }
}