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(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
解答:
class Solution {
public boolean exist(char[][] board, String word) {
boolean[][] visited = new boolean[board.length][board[0].length];//判断是否访问过
if(board.length * board[0].length < word.length()) return false;
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(backTrack(board, word, visited, i, j, 0)) return true;
}
}
return false;
}
public boolean backTrack(char[][] board, String word, boolean[][] visited, int i, int j, int index){//回溯法
if(index == word.length()) return true;//最后一个都匹配成功
if(i < 0 || i >= board.length || j < 0 || j >= board[0].length) return false;//条件不符
if(word.charAt(index) != board[i][j] || visited[i][j]) return false;
visited[i][j] = true;//以该点进行深度遍历
if(backTrack(board, word, visited, i + 1, j, index + 1) ||
backTrack(board, word, visited, i - 1, j, index + 1) ||
backTrack(board, word, visited, i, j + 1, index + 1) ||
backTrack(board, word, visited, i, j - 1, index + 1)) //上下左右是否成功
return true;
visited[i][j] = false;//该节点不符合,则回溯,修改访问为未访问
return false;
}
}