[leetcode] Word Search

Word Search

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.

 

思路：DFS算法。

注意点：由于每个字符只能匹配一次，所以匹配成功将该字符置为无用字符，比如‘#’等，然后继续匹配下一个字符。

 

class Solution
{
public:
  Solution(): isTrue(false) {}

private:
  void dfs(vector<vector<char> > &board,string word, int x, int y, int index)
  {
    if(index == word.size()) 
    {
      isTrue = true;
      return;
    }
    
    if(isTrue) return;
    if(x < 0 || x >= board.size()) return;
    if(y < 0 || y >= board[0].size()) return;
    if(board[x][y] != word[index]) return;

    board[x][y] = '#';
    dfs(board, word, x-1, y, index+1);
    dfs(board, word, x+1, y, index+1);
    dfs(board, word, x, y-1, index+1);
    dfs(board, word, x, y+1, index+1);
    board[x][y] = word[index];
  }
public:
  bool exist(vector<vector<char> > &board, string word)
  {
    if(board.empty() || board[0].empty()) return false;

    isTrue = false;
    int index = 0;
    for(int i=0; i<board.size(); i++)
    {
      for(int j=0; j<board[0].size(); j++)
      {
        if(!isTrue)
          dfs(board, word, i, j, index);
      }
    }

    return isTrue;
  }

private:
  bool isTrue;
};