請設計一個函數,用來判斷在一個矩陣中是否存在一條包含某字符串所有字符的路徑。路徑可以從矩陣中的任意一格開始,每一步可以在矩陣中向左、右、上、下移動一格。如果一條路徑經過了矩陣的某一格,那麼該路徑不能再次進入該格子。例如,在下面的3×4的矩陣中包含一條字符串“bfce”的路徑(路徑中的字母用加粗標出)。
[["a","b","c","e"]
,
["s","f","c","s"]
,
["a","d","e","e"]]
.
但矩陣中不包含字符串“abfb”的路徑,因爲字符串的第一個字符b佔據了矩陣中的第一行第二個格子之後,路徑不能再次進入這個格子。
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","d"]], word = "abcd"
輸出:false
提示:
1 <= board.length <= 200
1 <= board[i].length <= 200
來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof
著作權歸領釦網絡所有。商業轉載請聯繫官方授權,非商業轉載請註明出處。
分析:
深度優先搜索
解答:
class Solution {
public:
int x[4] = {0,0,-1,1};
int y[4] = {-1,1,0,0};
bool dps(vector<vector<char>> board, string word, int row, int col, int i, int j, int num){
if(num==word.size()) return true;
char tmp = board[i][j];
board[i][j] = '!';
for(int k=0;k<4;k++){
int coor_x = i + x[k];
int coor_y = j + y[k];
if(coor_x>=0&&coor_x<row&&coor_y>=0&&coor_y<col&&board[coor_x][coor_y]==word[num]){
if(dps(board, word, row, col, coor_x, coor_y, num+1)) return true;
}
}
board[i][j] = tmp;
return false;
}
bool exist(vector<vector<char>>& board, string word) {
int row = board.size();
int col = board[0].size();
for(int i=0;i<row;i++){
for(int j=0;j<col;j++){
if(board[i][j]==word[0]){
if(dps(board, word, row, col, i, j, 1)) return true;
}
}
}
return false;
}
};