[LeetCode] Word Search

[Problem]

Given a 2D board and a word, find if the word exists in thegrid.

The word can be constructed from letters of sequentiallyadjacent cell, where "adjacent" cells are those horizontally orvertically neighboring. The same letter cell may not be used morethan once.

For example,
Given board =

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

word = "ABCCED", -> returnstrue,
word = "SEE", -> returnstrue,
word = "ABCB", -> returnsfalse.

[Analysis]
深度优先搜索

[Solution]

class Solution {
public:
    // DFS
    bool DFS(vector<vector<char> > &board, int i, int j, int m, int n, string word, vector<vector<bool> > &visited){
        if(word.length() == 0){
            return true;
        }
        else if(i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[0] || visited[i][j] == true){
            return false;
        }
        else{
            visited[i][j] = true;
   
            // visit upper
            bool upper =  DFS(board, i-1, j, m, n, word.substr(1, word.length()-1), visited);
            if(upper){
                return true;
            }
   
            // visit down
            bool down = DFS(board, i+1, j, m, n, word.substr(1, word.length()-1), visited);
            if(down){
                return true;
            }
           
            // visit left
            bool left = DFS(board, i, j-1, m, n, word.substr(1, word.length()-1), visited);
            if(left){
                return true;
            }
   
            // visit right
            bool right = DFS(board, i, j+1, m, n, word.substr(1, word.length()-1), visited);
            if(right){
                return true;
            }
   
            visited[i][j] = false;
        }
       
        return false;
    }
   
    // exist
    bool exist(vector<vector<char> > &board, string word) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        // empty string
        if(word.length() == 0){
            return true;
        }
   
        // get number of rows
        int m = board.size();
        if(m == 0){
            return false;
        }
   
        // get number of columns
        int n = board[0].size();
   
        // create visited
            vector<vector<bool> > visited;
            for(int x = 0; x < m; ++x){
                vector<bool> row;
            for(int y = 0; y < n; ++y){
                row.push_back(false);
            }
            visited.push_back(row);
        }
   
        // DFS
        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j){
                if(board[i][j] == word[0]){
                    // DFS
                    bool contain = DFS(board, i, j, m, n, word, visited);
                    if(contain){
                        return true;
                    }
                }
            }
        }
        return false;
    }
};

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