扁平化网站下载win优化大师怎么样

题目描述

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

示例 2：

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

示例 3：

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

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

解决方案：

1、越界检查

2、终止条件：字符符合，字符串长度达到，该位置已遍历

3、单层循环逻辑：pos + 1，上下左右四方向判断

函数源码：

class Solution {
public:bool exist(vector<vector<char>>& board, string word) {if(board.empty())   return false;int m=board.size(),n=board[0].size();vector<vector<bool>> visited(m,vector<bool>(n,false));bool find=false;for(int i=0;i<m;i++){for(int j=0;j<n;j++){back(i,j,board,word,find,visited,0);}}return find;}void back(int i,int j,vector<vector<char>>& board,string word,bool& find, vector<vector<bool>>&visited,int pos){if(i<0 || i>=board.size() || j<0 || j>=board[0].size())     return;if(visited[i][j] || find || board[i][j] != word[pos])       return;if(pos == word.size()-1){find=true;return;}visited[i][j]=true; //已遍历back(i+1,j,board,word,find,visited,pos+1);back(i-1,j,board,word,find,visited,pos+1);back(i,j-1,board,word,find,visited,pos+1);back(i,j+1,board,word,find,visited,pos+1);visited[i][j] = false;}};