1. 题目
请设计一个函数,用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始,每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格,那么该路径不能再次进入该格子。例如,在下面的3×4的矩阵中包含一条字符串“bfce”的路径。
[["a","b","c","e"],
["s","f","c","s"],
["a","d","e","e"]]
但矩阵中不包含字符串“abfb”的路径,因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后,路径不能再次进入这个格子。
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","d"]], word = "abcd"
输出:false
提示:
1 <= board.length <= 200
1 <= board[i].length <= 200
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof
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2. 题目
相同题目:
LeetCode 79. 单词搜索(回溯DFS)
class Solution {
bool found = false;
vector<vector<int>> dir = {{1,0},{0,1},{0,-1},{-1,0}};
int m, n;
public:
bool exist(vector<vector<char>>& board, string word) {
int i, j;
m = board.size(), n = board[0].size();
for(i = 0; i < board.size(); ++i)
{
for(j = 0; j < board[0].size(); ++j)
if(!found)
dfs(board,i,j,word,0);
else
return found;
}
return found;
}
void dfs(vector<vector<char>>& board, int i, int j, string& word, int w)
{
if(found || (w<word.size() && board[i][j] != word[w]))
return;
if(w==word.size()-1)
{
found = true;
return;
}
int x, y;
char ch;
for(int k = 0; k < 4; ++k)
{ //4个方向
x = i+dir[k][0];
y = j+dir[k][1];
if(x >=0 && x<m && y>=0 && y<n)
{
ch = board[i][j];//当前位置记录下来
board[i][j] = '*';//标记走过了
dfs(board,x,y,word,w+1);
board[i][j] = ch;//回溯,还原现场
}
}
}
};