[LeetCode]Word Search 回溯

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.

Hide Tags

Array Backtracking

 

---

 

    这事一道回溯题，写的有点重复，因为没有将多个if 合在一起。

 

 

 

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        if(word.length()<) return true;
        if(board.size()==||board[].size()==) return false;
        for(int i =;i<board.size();i++){
            for(int j =;j<board[].size();j++){
                if(board[i][j]==word[]&&helpFun(board,word,,i,j))
                    return true;
            }
        }
        return false;
    }

    bool helpFun(vector<vector<char> >&board,string & word,int idx,int beg_i,int beg_j)
    {
        if(idx == word.size())  return true;
        char tmp = board[beg_i][beg_j];
        board[beg_i][beg_j] = '*';
        if(beg_i>&&board[beg_i-][beg_j]==word[idx]&&helpFun(board,word,idx+,beg_i-,beg_j)){
            board[beg_i][beg_j] = tmp;
            return true;
        }
        if(beg_i<board.size()-&&board[beg_i+][beg_j]==word[idx]&&helpFun(board,word,idx+,beg_i+,beg_j)){
            board[beg_i][beg_j] = tmp;
            return true;
        }
        if(beg_j>&&board[beg_i][beg_j-]==word[idx]&&helpFun(board,word,idx+,beg_i,beg_j-)){
            board[beg_i][beg_j] = tmp;
            return true;
        }
        if(beg_j<board[].size()-&&board[beg_i][beg_j+]==word[idx]&&helpFun(board,word,idx+,beg_i,beg_j+)){
            board[beg_i][beg_j] = tmp;
            return true;
        }
        board[beg_i][beg_j] = tmp;
        return false;
    }
};

int main()
{
    vector<vector< char> > board{{'A','B','C','E'},{'S','F','C','S'},{'A','D','E','E'}};
    Solution sol;
    cout<<sol.exist(board,"ABCB")<<endl;
    return ;
}