79. Word Search 回溯算法_word回溯-优快云博客

本文详细介绍了二维网格中查找单词的算法实现，通过回溯法解决此问题，包括设置索引避免重复使用字母格子及使用数组记录已访问位置。通过实例演示算法的应用，包括边界条件的处理及递归终止条件。

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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);}
        // 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;
    }
};