leetcode-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

.

class Solution {
public:

bool vis[1000][1000];
int  row,length;
bool backtracing(vector<vector<char> >&grid,string &word,int cur,int i,int j){
if(cur==length) //递归终止条件，能走到这一步，说明已经匹配完了。
return true;
if(j>=0&&i>=0&&j<grid[i].size()&&i<row&&!vis[i][j]&&grid[i][j]==word[cur]){

vis[i][j]=true;//避免走回头路，做个标记说明
if(backtracing(grid,word,cur+1,i,j+1))//向右
return true;
if(backtracing(grid,word,cur+1,i+1,j))//向下
return true;
if(backtracing(grid,word,cur+1,i-1,j))//向上
return true;
if(backtracing(grid,word,cur+1,i,j-1))//向左
return true;
vis[i][j]=false;//记得还原
}
return false;
}
bool exist(vector<vector<char> > &board, string word) {
row=board.size();
length=word.length();
for(int i=0;i<row;++i){
int cal=board[i].size();//因为每个单词不等长，所以要计算
for(int j=0;j<cal;++j){
if(word[0]==board[i][j]){//先确定递归入口
if(backtracing(board,word,0,i,j))
return true;
}
}
}
return false;
}
};