79. 单词搜索

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：


输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true

思路：

和岛屿的思路基本相同，只不过这里在进入递归之前进行剪枝，减少了栈的空间消耗

class Solution {
public:
    bool dfs(vector<vector<char>>& board,int i,int j,string word,int start){
        if( board[i][j] != word[start])   return false;
        if(start == word.size() - 1)    return true;
        char c =board[i][j];
        board[i][j] = '.';
        for(int index = 0; index < 4; ++index){
            int next_i = i + di[index];
            int next_j = j + dj[index];
            if(next_i < 0 || next_j < 0 || next_i >= board.size() || next_j >= board[0].size()) continue;
            if(dfs(board, next_i,next_j,word,start+1)) return true;
        }
        board[i][j] = c;
        return false;
    }
    bool exist(vector<vector<char>>& board, string word) {
        int count = 0;
        //遍历board 如果符合 则进入dfs的下一层
        for(int i = 0;i < board.size();i++){
            for(int j = 0; j < board[i].size();j++){
                    if(dfs(board,i,j,word,0)){
                        return true;
                    }
            }
        }
        return false;
    }
private:
    int di[4] = {-1,0,1,0};
    int dj[4] = {0,1,0,-1};
};