Leetcode: Word search

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,
GivenBoard=

[  ["ABCE"],  ["SFCS"],  ["ADEE"]]

Word="ABCCED",-> Returnstrue,
Word="SEE",-> Returnstrue,
Word="ABCB",-> Returnsfalse.

Address: https://oj.leetcode.com/problems/word-search/

Algorithm: Use the DFS plus backtracing method. First, start searching from every position on the board. Assume that the current traversal is to the position (I, j). If the first character of word is equal to that of the Board [I] [J], set the traversal position to true, and then call the function existcore to perform DFS. The last three parameters of the existcore function are represented. Currently, the traversal is to the position (I, j, and wait for the second POS character of the word to match. In the existcore function, if the last word of the word has been matched, the match is completed. Otherwise, the traversal starts from the (I, j) position and goes up to the left and right positions, if conditions are met, existcore is recursively called. Note: If the solution cannot be found in a certain direction, the corresponding flag position should be cleared and traced back to the original position. Code:

 1 class Solution { 2 public: 3     bool exist(vector<vector<char> > &board, string word) { 4         if(board.empty())   return false; 5         if(word.empty())    return true; 6         int m = board.size(); 7         vector<vector<bool> > flag; 8         for(int i = 0; i < m; ++i){ 9             int n = board[i].size();10             flag.push_back(vector<bool>(n,false));11         }12         for(int i = 0; i < m; ++i){13             int n = board[i].size();14             for(int j = 0; j < n; ++j){15                 if(word[0] == board[i][j]){16                     flag[i][j] = true;17                     if(existCore(board,word,flag,i,j,1))18                         return true;19                     flag[i][j] = false;20                 }21             }22         }23         return false;24     }25     bool existCore(vector<vector<char> > &board, string &word, vector<vector<bool> > &flag, int i, int j, int pos){26         if(pos == word.size()){27             return true;28         }29         if(i > 0 && j < board[i-1].size() && !flag[i-1][j] && board[i-1][j] == word[pos]){30             flag[i-1][j] = true;31             if(existCore(board,word,flag,i-1,j,pos+1)){32                 return true;33             }34             flag[i-1][j] = false;35         }36         if(j < board[i].size() - 1 && !flag[i][j+1] && board[i][j+1] == word[pos]){37             flag[i][j+1] = true;38             if(existCore(board,word,flag,i,j+1,pos+1)){39                 return true;40             }41             flag[i][j+1] = false;42         }43         if(i < board.size() - 1 && j < board[i+1].size() && !flag[i+1][j] && board[i+1][j] == word[pos]){44             flag[i+1][j] = true;45             if(existCore(board,word,flag,i+1,j,pos+1)){46                 return true;47             }48             flag[i+1][j] = false;49         }50         if(j > 0 && !flag[i][j-1] && board[i][j-1] == word[pos]){51             flag[i][j-1] = true;52             if(existCore(board,word,flag,i,j-1,pos+1)){53                 return true;54             }55             flag[i][j-1] = false;56         }57         return false;58     }59 };

 

Leetcode: Word search