《剑指offer》面试题12. 矩阵中的路径
问题描述
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（路径中的字母用加粗标出）。
[["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
代码
使用额外空间：
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int length = word.size();
int m = board.size(),n = board[0].size();
vector<vector<bool>> flag(m,vector<bool>(n,false));
for(int i = 0; i < m; ++i)
{
for(int j = 0; j < n; ++j)
if( check(board,i,j,word,0,flag))
return true;;
}
return false;
}
bool check(vector<vector<char>>& board, int indx,int indy,string word,int ind,vector<vector<bool>> &flag)
{
if(ind == word.size())return true;
int m = board.size(),n = board[0].size();
if(indx < 0 || indx >= m || indy < 0 || indy >= n || board[indx][indy]!=word[ind]||flag[indx][indy])
return false;
flag[indx][indy] = true;
bool res = check(board,indx+1,indy,word,ind+1,flag)||
check(board,indx,indy+1,word,ind+1,flag)||
check(board,indx-1,indy,word,ind+1,flag)||
check(board,indx,indy-1,word,ind+1,flag);
flag[indx][indy] = false;
return res;
}
};
结果：
执行用时 :448 ms, 在所有 C++ 提交中击败了14.88%的用户
内存消耗 :187.6 MB, 在所有 C++ 提交中击败了100.00%的用户
代码
不使用额外空间
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int length = word.size();
int m = board.size(),n = board[0].size();
for(int i = 0; i < m; ++i)
{
for(int j = 0; j < n; ++j)
if( check(board,i,j,word,0))
return true;;
}
return false;
}
bool check(vector<vector<char>>& board, int indx,int indy,string word,int ind)
{
if(ind == word.size())return true;
int m = board.size(),n = board[0].size();
if(indx < 0 || indx >= m || indy < 0 || indy >= n || board[indx][indy]!=word[ind]||board[indx][indy]=='*')
return false;
char c = board[indx][indy];
board[indx][indy]='*';
bool res = check(board,indx+1,indy,word,ind+1)||
check(board,indx,indy+1,word,ind+1)||
check(board,indx-1,indy,word,ind+1)||
check(board,indx,indy-1,word,ind+1);
board[indx][indy] = c;
return res;
}
};
结果：
执行用时 :392 ms, 在所有 C++ 提交中击败了24.83%的用户
内存消耗 :187 MB, 在所有 C++ 提交中击败了100.00%的用户
