79. Word Search 回溯算法

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

.

分析：

回溯算法的典型应用，需要注意的有两点。

1，设置index保证找到一个后及时退出当层循环。

2。设置一个数组mm，记录跑过的位置。

代码：

static int mm[100][100];

class Solution {

public:

bool find(int i,int j,int t,int m, int n,vector<vector<char>> board, string word)

{

    

    if(t==word.size()-1) return true;

    else

    {

        // cout<<i<<j<<endl;

        bool index=false;

        mm[i][j]=1;

        if(!index && (0<=(i+1))&&(i+1)<m&& (0<=j<n )&& (board[i+1][j]==word[t+1])&&mm[i+1][j]==0)

        { index=find(i+1,j,t+1,m,n,board,word);}

        if(!index && 0<=i &&i<m &&(0<=(j+1))&&(j+1)<n && (board[i][j+1]==word[t+1])&&mm[i][j+1]==0)

        {index=find(i,j+1,t+1,m,n,board,word);}

        if(!index && (0<=(i-1))&&(i-1)<m && (0<=j<n )&& (board[i-1][j]==word[t+1])&&mm[i-1][j]==0)

        { index=find(i-1,j,t+1,m,n,board,word);}

        if(!index && (0<=i)&&i<m && (0<=(j-1))&&(j-1)<n && (board[i][j-1]==word[t+1])&&mm[i][j-1]==0)

        {index=find(i,j-1,t+1,m,n,board,word);}

        // if(!index && (0<=(i+1))&&((i+1)<m )&& (0<=j)&&j<n && (board[i+1][j]==word[t+1])&&mm[i+1][j]==0)

        // { index=find(i+1,j,t+1,m,n,board,word);}

        // if(!index && (0<=i) &&i<m &&(0<=(j+1))&&(j+1)<n && (board[i][j+1]==word[t+1])&&mm[i][j+1]==0)

        // {index=find(i,j+1,t+1,m,n,board,word)
4000
;}

        // if(!index && (0<=(i-1))&&(i-1)<m && (0<=j)&&j<n && (board[i-1][j]==word[t+1])&&mm[i-1][j]==0)

        // { index=find(i-1,j,t+1,m,n,board,word);}

        // if(!index && (0<=i) &&i<m && (0<=(j-1))&&(j-1)<n && (board[i][j-1]==word[t+1])&&mm[i][j-1]==0)

        // {index=find(i,j-1,t+1,m,n,board,word);}

        mm[i][j]=0;

        return index;

    }

}

    bool exist(vector<vector<char>>& board, string word) {

        if(word.size()==0 || board.size()==0) return false;

        int m=board.size();

        int n=board[0].size();

        for(int i=0;i<board.size();++i)

        for(int j=0;j<board[0].size();++j)

        {

            if(board[i][j]==word[0])

            {

 

                if(find(i,j,0,m,n,board,word)) return true;

            }

        }

        return false;

    }

};