leetcode 079 —— 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： 思路，用一个矩阵记录被使用的单词，但是代码跑了44ms，还有优化的空间

class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
bool judge = false;
int level = 0;
vector<vector<bool>> record(board.size(), vector<bool>(board[0].size(), false));
for (int m = 0; m <board.size(); m++){
for (int n = 0; n <board[0].size(); n++){
if (board[m]
== word[0]){
record[m]
= true;
level = 1;
scan(m, n, judge, level, word, record, board);
if (judge) return judge;
record[m]
= false;
}
}
}
return judge;
}
void scan(int i, int j, bool &judge,int &level, string &word,vector<vector<bool>> &record,vector<vector<char>> &board){
if (level== word.size()){
judge = true;
return;
}
if (i + 1 < board.size() && !record[i + 1][j] && board[i + 1][j] == word[level]){ //往右
record[i + 1][j] = true;
level++;
scan(i + 1, j, judge, level, word, record, board);
level--;
if (judge) return;
record[i + 1][j] = false;
}
if (i - 1 >= 0 && !record[i - 1][j] && board[i - 1][j] == word[level]){ //往左
record[i - 1][j] = true;
level++;
scan(i - 1, j, judge, level, word, record, board);
level--;
if (judge) return;
record[i - 1][j] = false;
}
if (j + 1 < board[0].size() && !record[i][j + 1] && board[i][j+1] == word[level]){ //往下
record[i][j+1] = true;
level++;
scan(i, j + 1, judge, level, word, record, board);
level--;
if (judge) return;
record[i][j+1] = false;
}
if (j - 1 >= 0 && !record[i][j - 1] && board[i][j-1] == word[level]){ //往上
record[i][j-1] = true;
level++;
scan(i , j-1, judge, level, word, record, board);
if (judge) return;
level--;
record[i][j-1] = false;
}
}
};