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.
Example:
board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] Given word = "ABCCED", return true. Given word = "SEE", return true. Given word = "ABCB", return false.
解題思路：
遞歸法：用cur記錄當前遍歷word位置，當最終cur == word的長度，說明含有此單詞，對每個(i,j)位置，若與word的cur處的字母不一致或者該位置已經遍歷過（用visit數組標記），則返回錯誤。若一致，則遍歷（i，j）的周圍的字母與word的cur+1處的字符。
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if((board.size() == 0 || board[0].size() == 0) && word.length() > 0)
return false;
else if(word.length() == 0)
return true;
m = board.size(); n = board[0].size();
vector<vector<bool>>visit(m, vector<bool>(n, false));
for(int i = 0; i < m; ++ i)
for(int j = 0; j < n; ++ j){
if(dfs(board, visit, i, j, word, 0))
return true;
}
return false;
}
private:
int m, n;
bool dfs(vector<vector<char>>& board, vector<vector<bool>>& visit, int row, int col, string word, int cur){
if(cur == word.length())
return true;
if(row < 0 || row == m || col < 0 || col == n)
return false;
if(visit[row][col] == true || board[row][col] != word[cur])
return false;
visit[row][col] = true;
bool flag = dfs(board, visit, row+1, col, word, cur+1) || dfs(board, visit, row-1, col, word, cur+1) || dfs(board, visit, row, col+1, word, cur+1) || dfs(board, visit, row, col-1, word, cur+1);
visit[row][col] = false;
return flag;
}
};