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：深度优先搜索。注意每条路径搜索完之后，所有的 visited 重置为 false (未访问).
typedef pair<int, int> point;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
bool dfs(vector<vector<char> > &board, string& word, int id, point p, vector<vector<bool> > &visited) {
if(id == word.size()) return true;
visited[p.first][p.second] = true;
for(int i = 0; i < 4; ++i) {
int x = p.first + dx[i], y = p.second +dy[i];
if(x < 0 || x >= board.size() || y < 0 || y >= board[0].size() || visited[x][y]) continue;
if(board[x][y] == word[id] && dfs(board, word, id+1, point(x, y), visited))
return true;
visited[x][y] = false;
}
return false;
}
class Solution {
public:
bool exist(vector<vector<char> > &board, string word) {
if(word == "") return true;
if(board.size() == 0 || board[0].size() == 0) return false;
int row = board.size(), col = board[0].size();
vector<vector<bool> > visited(row, vector<bool>(col, 0));
for(int r = 0; r < row; ++r) {
for(int c = 0; c < col; ++c) {
if(board[r][c] == word[0] && dfs(board, word, 1, point(r,c), visited))
return true;
visited[r][c] = false;
}
}
return false;
}
};
