[leetcode-79-Word Search]

Posted 2020-09-13 hellowOOOrld

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.

思路：

深度优先遍历。

bool dfs(vector< vector< char> >& board, string word,int row ,int col ,int start,int m,int n,int length)
{
    char curChar = board[row][col];
    bool res = false;

    if(curChar != word[start]) return false;
    if(start == length - 1) return true;

    board[row][col] = ‘*‘;
    if(row>0) res = dfs(board,word,row-1,col,start+1,m,n,length);
    if(!res && row < m-1) res = dfs(board,word,row+1,col,start+1,m,n,length);
    if(!res && col>0) res = dfs(board,word,row,col-1,start+1,m,n,length);
    if(!res && col<n-1) res = dfs(board,word,row,col+1,start+1,m,n,length);
    board[row][col] = curChar;

    return res;
}
bool exist(vector< vector< char> >& board, string word)
{
    int m = board.size() , n = board[0].size(), length = word.size();

    if( m && n && length)
    {
        for(int i = 0;i < m;i++)
        {
            for(int j =0;j<n;j++)
            {
                if(dfs(board,word,i,j,0,m,n,length))
                    return true;
            }
        }
    }
    return false;
}