LeetCode 79: Word Search

Difficulty: 3

Frequency: 4

Problem:

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.

Solution:

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (word.size()==0)            return true;                if (word.size()==1)        {            for (int i = 0; i<board.size(); i++)            {                for (int j = 0; j<board[i].size(); j++)                {                    if (board[i][j]==word[0])                        return true;                }            }            return false;        }                vector<vector<char> > visited(board.size());        for (int i = 0; i<visited.size(); i++)        {            visited[i].resize(board[i].size(), 0);        }        for (int i = 0; i<board.size(); i++)        {            for (int j = 0; j<board[i].size(); j++)            {                if (board[i][j]==word[0]&&DFS(board, visited, i, j, word, 1))                    return true;            }        }        return false;    }    bool DFS (vector<vector<char> > & board, vector<vector<char> > & visited, int row, int col, const string & word, int i_word)    {        visited[row][col] = 1;        if (row!=0&&!visited[row-1][col])        {            if (board[row-1][col]==word[i_word])            {                if (i_word==word.size()-1)                    return true;                else if (DFS(board, visited, row-1, col, word, i_word+1))                    return true;            }                        }        if (row<visited.size()-1&&!visited[row+1][col])        {            if (board[row+1][col]==word[i_word])            {                if (i_word==word.size()-1)                    return true;                else if (DFS(board, visited, row+1, col, word, i_word+1))                    return true;            }        }        if (col!=0&&!visited[row][col-1])        {            if (board[row][col-1]==word[i_word])            {                if (i_word==word.size()-1)                    return true;                else if (DFS(board, visited, row, col-1, word, i_word+1))                    return true;            }        }        if (col<visited[row].size()-1&&!visited[row][col+1])        {            if (board[row][col+1]==word[i_word])            {                if (i_word==word.size()-1)                    return true;                else if (DFS(board, visited, row, col+1, word, i_word+1))                    return true;            }        }                visited[row][col] = 0;        return false;    }};

Notes:

DFS.