題目描述

給定一個?m x n?二維字符網格?board?和一個字符串單詞?word?。如果?word?存在于網格中，返回?true?；否則，返回?false?。

單詞必須按照字母順序，通過相鄰的單元格內的字母構成，其中“相鄰”單元格是那些水平相鄰或垂直相鄰的單元格。同一個單元格內的字母不允許被重復使用。

示例 1：

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

示例 2：

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

示例 3：

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

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board?和?word?僅由大小寫英文字母組成

解決方案：

1、越界檢查

2、終止條件：字符符合，字符串長度達到，該位置已遍歷

3、單層循環邏輯：pos + 1，上下左右四方向判斷

函數源碼：

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; //已遍歷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;}};