题目描述(Medium)
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.
题目链接
https://leetcode.com/problems/word-search/description/
Example 1:
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.
算法分析
注意几个终止条件，越界、已访问、不相等。
提交代码：
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if (board.empty() || board[0].empty()) return false;
const int m = board.size();
const int n = board[0].size();
vector<vector<bool>> visited(m, vector<bool>(n, false));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if(DFSWordSearch(board, visited, word, 0, i, j))
return true;
}
}
return false;
}
private:
bool DFSWordSearch(vector<vector<char>>& board, vector<vector<bool>>& visited,
string& word, int index, int x, int y) {
if (index == word.size()) return true;
if (x < 0 || y < 0 || x >= board.size() || y >= board[0].size()) return false;
if (visited[x][y]) return false;
if (word[index] != board[x][y]) return false;
visited[x][y] = true;
bool result = DFSWordSearch(board, visited, word, index + 1, x + 1, y)
|| DFSWordSearch(board, visited, word, index + 1, x - 1, y)
|| DFSWordSearch(board, visited, word, index + 1, x, y + 1)
|| DFSWordSearch(board, visited, word, index + 1, x, y - 1);
visited[x][y] = false;
return result;
}
};