题目:
给定一个二维网格和一个单词,找出该单词是否存在于网格中。
单词必须按照字母顺序,通过相邻的单元格内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
示例:
board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.
思路:
暴力求解的法子,从每个字符开始进行深度优先搜索寻找匹配的字符串。
代码:
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if (board.size() == 0) return false;
for (int i = 0; i < board.size();i++)
{
for (int j = 0; j < board[0].size(); j++)
{
if (dfs(board, word, i, j, 0))
return true;
}
}
return false;
}
bool dfs(vector<vector<char>>& board, string word, int i, int j, int length)
{
if(i>=board.size()||j>=board[0].size()||length>=word.length()||i<0||j<0||board[i][j]!=word[length])
{
return false;
}
if (length == (word.size() - 1) && board[i][j] == word[length])
return true;
char temp = board[i][j];
board[i][j] = '0';
bool flag = dfs(board, word, i + 1, j, length + 1) || dfs(board, word, i, j - 1, length + 1) || dfs(board, word, i, j + 1, length + 1) || dfs(board, word, i - 1, j, length + 1);
board[i][j] = temp;
return flag;
}
};