backtracking vs DFS

还在理解中。。。

leetcode 200. Number of Islands

Input:
11110
11010
11000
00000

Output: 1

vs
leetcode 79. Word Search

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

两题写法非常相似，怎么理解？

The island solution is using DFS:

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int res = 0;
        for (int i = 0; i < grid.size(); ++i)
            for (int j = 0; j < grid[0].size(); ++j)
                if ('1' == grid[i][j])
                {
                    res++;
                    DFS(grid, i, j);
                }
        return res;
    }
private:
    void DFS(vector<vector<char>>&grid, int i, int j)
    {
        if(grid[i][j] == '1')
        {
            grid[i][j] = '0';
        }
        
        if(i>0 && grid[i-1][j] == '1') DFS(grid,i-1,j);
        if(i+1 < grid.size() && grid[i+1][j] == '1') DFS(grid,i+1,j);
        if(j>0 && grid[i][j-1] == '1') DFS(grid,i,j-1);
        if(j+1 < grid[0].size() && grid[i][j+1] == '1') DFS(grid,i,j+1);
    }
};

The word search solution is using backtracking:

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