标签：private 不包含 sla 进入 tps null solution ted 矩阵
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 <= 2001 <= board[i].length <= 200class Solution { public boolean exist(char[][] board, String word) { if(word == null || word.length() == 0) return true; // List<Character> list = new ArrayList<>(); boolean[][] visited = new boolean[board.length][board[0].length]; for(int i = 0;i < board.length;i++){ for(int j = 0;j < board[0].length;j++){ if(dfs(board,i,j,0,word,visited)) return true; } } return false; } private boolean dfs(char[][] board,int i, int j,int index,String word,boolean[][] visited){ if(index == word.length()) return true; if(i >= board.length || i < 0 || j < 0 || j >= board[0].length || visited[i][j] || index >= word.length() || board[i][j] != word.charAt(index)){ return false; } visited[i][j] = true; // list.add(board[i][j]); if(dfs(board,i + 1,j,index + 1,word,visited) || dfs(board,i,j + 1,index + 1,word,visited) || dfs(board,i - 1,j,index + 1,word,visited) ||dfs(board,i,j - 1,index + 1,word,visited) ){ return true; } // list.remove(list.size() - 1); visited[i][j] = false; return false; } }
标签：private 不包含 sla 进入 tps null solution ted 矩阵
原文地址：https://www.cnblogs.com/zzytxl/p/12637131.html
