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.
Example:
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.
思路 ： DFS ， 写得思维有点混乱了
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int n = board.size();
int m = board[0].size();
int len ;
m_word = word; //保存字符串
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < m; ++j)
{
if (board[i][j] == word[0])
{
len = 1;
vector<vector<bool>> IsVisit(n, vector<bool>(m, false)); //初始化一个bool数组来判断是否访问过
IsVisit[i][j] = true;
if (Search(board,IsVisit , i, j, len)) return true;
IsVisit.clear();
}
}
}
return false;
}
bool Search(vector<vector<char>>&board, vector<vector<bool>> & IsVisit ,int xnow, int ynow, int len )
{
if (len == m_word.size()) return true; //已经匹配长度 len-1
int x, y;
int direction[4][2] = { { 0, 1 }, { 1, 0 }, { 0, -1 }, {-1,0} }; //顺时钟
for (int i = 0; i <= 3; ++i)
{
x = xnow + direction[i][0];
y = ynow + direction[i][1];
if (x >= 0 && x < board.size() && y >= 0 && y < board[0].size() && !IsVisit[x][y] && board[x][y] == m_word[len])
{
len ++;
IsVisit[x][y] = true;
if (Search(board, IsVisit, x, y, len)) return true;
else
{
len --;
IsVisit[x][y] = false;
}
}
}
return false;
}
private :
string m_word;
};
int main()
{
vector<vector<char>> hah = { { 'A', 'B', 'C', 'E' }, { 'S', 'F', 'C', 'S' }, { 'A', 'D', 'E', 'E' } };
Solution a;
cout << a.exist(hah, "ABCCED");
system("pause");
return 0;
}