[leetcode] 79. Word Search 解题报告

题目链接: https://leetcode.com/problems/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 =

[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

word = 

"ABCCED"

,
-> returns 

true

,
word = 

"SEE"

,
-> returns 

true

,
word = 

"ABCB"

,
-> returns 

false

.

思路: 依然是个DFS, 需要注意的是要标记访问过的位置, 并且当搜索完以后返回要将其标记设置为未访问.
代码如下:

class Solution {
public:
bool DFS(vector<vector<char>>& board, vector<vector<bool>>& hash, string& word, int y, int x, int k)
{
if(k >= word.size()) return true;
int m = board.size(), n = board[0].size();
if(y<0 ||y>=m ||x<0 ||x>=n||hash[y][x]||board[y][x]!=word[k]) return false;
hash[y][x] = true;
if(DFS(board, hash, word, y+1, x, k+1)) return true;
if(DFS(board, hash, word, y-1, x, k+1)) return true;
if(DFS(board, hash, word, y, x+1, k+1)) return true;
if(DFS(board, hash, word, y, x-1, k+1)) return true;
hash[y][x] = false;
return false;
}

bool exist(vector<vector<char>>& board, string word) {
if(board.size() ==0) return false;
int m = board.size(), n = board[0].size();
vector<vector<bool>> hash(m, vector<bool>(n, false));
for(int i =0; i< m; i++)
for(int j =0; j<n; j++)
if(DFS(board, hash, word, i, j, 0)) return true;
return false;
}

};