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,
Given board =

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

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        if(word.size() == 0)            return true;        for(int i = 0; i < board.size(); i++)            for(int j = 0; j < board[i].size(); j++)        {            if(dfs(0, board, word, i, j))                return true;        }        return false;    }    bool dfs(int index, vector<vector<char> > &board, string word, int x, int y)    {        if(board[x][y] != word[index])            return false;        else        {            if(index == word.size() - 1)                return true;            char store = board[x][y];            if(x-1 >= 0)            {                board[x][y] = '#';                if(dfs(index+1, board, word, x-1, y))                    return true;                board[x][y] = store;            }            if(x+1 < board.size())            {                board[x][y] = '#';                if(dfs(index+1, board, word, x+1, y))                    return true;                board[x][y] = store;            }            if(y-1 >= 0)            {                board[x][y] = '#';                if(dfs(index+1, board, word, x, y-1))                    return true;                board[x][y] = store;            }            if(y+1 < board[0].size())            {                board[x][y] = '#';                if(dfs(index+1, board, word, x, y+1))                    return true;                board[x][y] = store;            }            return false;        }                   }};

Round 2:

class Solution {public:    bool exist(vector<vector<char> > &board, string word) {        if(board.size() == 0)            return false;        int sizeX = board.size();        int sizeY = board[0].size();        for(int i = 0; i < sizeX; i++)            for(int j = 0; j < sizeY; j++)            {                if(dfs(i, j, word, board, 0))                    return true;            }        return false;    }private:    bool dfs(int x, int y, string target, vector<vector<char> > &board, int index)    {        if(board[x][y] != target[index])            return false;        if(index == target.size()-1)            return true;        if(x-1 >= 0 && board[x-1][y] == target[index+1])        {            char temp = board[x][y];            board[x][y] = '#';            if(dfs(x-1, y, target, board, index+1))            {                return true;            }            board[x][y] = temp;        }        if(y-1 >= 0 && board[x][y-1] == target[index+1])        {            char temp = board[x][y];            board[x][y] = '#';            if(dfs(x, y-1, target, board, index+1))            {                return true;            }            board[x][y] = temp;        }        if(y+1 < board[0].size() && board[x][y+1] == target[index+1])        {            char temp = board[x][y];            board[x][y] = '#';            if(dfs(x, y+1, target, board, index+1))            {                return true;            }            board[x][y] = temp;        }        if(x+1 < board.size() && board[x+1][y] == target[index+1])        {            char temp = board[x][y];            board[x][y] = '#';            if(dfs(x+1, y, target, board, index+1))            {                return true;            }            board[x][y] = temp;        }        return false;                    }};