【LeetCode】Word Search

题目描述：

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

.

这题目做的有点蛋疼……一眼看去用DFS肯定可以解出这道题的，但按照之前做题的经验，感觉应该不会这么简单，于是在寻找能优化到O（n）的方法。想法是用矩阵s[i][j]来储存board[i][j]元素位于word[k]位置时是否遍历过，但是提交后发现有种情况是无法通过的，例如test
case：

Input:	["ABCE","SFES","ADEE"], "ABCESEEEFS"
Output:	false
Expected:	true

A→B→C→E

↓

S←F←E S

↑ ↓

A D E←E

正确的顺序应该是按照箭头所示，但如果到C时先向下走，再向右走，这样虽然无法得出正确结果，但是S已经遍历过了，再从C向右向下走的时候就会返回false，因此这种方法是行不通的。

最后无奈之下直接用DFS提交了下，居然过了……过了……

以后果然还是应该先写一个出来再优化……orz

代码如下：

class Solution {
public:
bool exist(vector<vector<char> > &board, string word) {
if (!word.length() || board.empty())
return false;
vector<vector<vector<int>>> dp(board.size(), vector<vector<int>>(board[0].size(), vector<int>()));
for (int i = 0; i < board.size();i++)
for (int j = 0; j < board[i].size(); j++){
if (board[i][j] == word[0] && isExt(board, word, dp, 0, i, j))
return true;
}
return false;
}
bool isExt(vector<vector<char>> &board, string &word, vector<vector<vector<int>>> &dp, int windex, int i, int j){
char c = word[windex];
if (board[i][j] != c || windex >= word.length())
return false;
if (windex == word.length() - 1)
return true;
vector<int> searched = dp[i][j];
for (int k = 0; k < searched.size();k++)
if (searched[k] == windex)
return false;
dp[i][j].push_back(windex);
windex++;
c = board[i][j];
board[i][j] = 0;
if (i - 1 >= 0 && isExt(board, word, dp, windex, i - 1, j)) return true;
if (i + 1 < board.size() && isExt(board, word, dp, windex, i + 1, j)) return true;
if (j - 1 >= 0 && isExt(board, word, dp, windex, i, j - 1)) return true;
if (j + 1 < board[0].size() && isExt(board, word, dp, windex, i, j + 1)) return true;
board[i][j] = c;
return false;
}
};