leetcode79- 找单词

dfs 广度搜索： 额 说白了就是每次往 上下左右四周找匹配的·字符罢了

同时还要注意当前字符是否已经使用过了。

 

#include<vector>
#include<string>
#include<iostream>
using namespace std;
class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
    	if(board.size()==0 || board[0].size()==0){
    		return false;
    	}
        int r = board.size();
    	int c = board[0].size();
    	
    	for(int i=0;i<r;++i){
    		for(int j=0;j<c;++j){
    			bool result = findWord(board, 0, word, i, j);
    			if(result){
    				return true;
    			}
    		}
    	}
        return false;
    }
    bool findWord(vector<vector<char>>& board, int k, string& word, int row,int col){
    	int rows = board.size()-1;
    	int cols = board[0].size()-1;
    	// 当·长度足够 即代表找到了
    	if(word.size() == k){
    		return true;
    	}
    	if(row < 0 || row > rows || col < 0 || col > cols||board[row][col] != word[k]){
    		return false;
    	}
    	char help = board[row][col];
    	board[row][col] = '*';
    	bool res = findWord(board,k+1, word, row+1, col)||
    	findWord(board, k+1, word, row-1, col)||
    	findWord(board, k+1, word, row, col+1)||
    	findWord(board, k+1, word, row, col-1);
    	board[row][col] = help;
    	return res;
    }
};
int main(){
	vector<vector<char>> board({{'C','A','A'},{'A','A','A'},{'B','C','D'}});
	// vector<vector<char>> board({{'A'}});
	Solution s;
	cout << s.exist(board, "AAB");

	return 0;
}