剑指 Offer 12. 矩阵中的路径
推荐写法
把判断条件都写在dfs函数开头（对节点进行处理，尽量不要对边进行处理）
写法一
class Solution {boolean[][] vis;public boolean exist(char[][] board, String word) {int m = board.length, n = board[0].length;vis = new boolean[m][n];for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){if(dfs(board, i, j, word, 0)) return true;}}return false;}boolean dfs(char[][] board, int x, int y, String word, int ind){if(x < 0 || x >= board.length || y < 0 || y >= board[0].length || vis[x][y] || board[x][y] != word.charAt(ind)) return false;if(ind == word.length() - 1) return true;vis[x][y] = true;if(dfs(board, x + 1, y, word, ind + 1)) return true;if(dfs(board, x - 1, y, word, ind + 1)) return true;if(dfs(board, x, y + 1, word, ind + 1)) return true;if(dfs(board, x, y - 1, word, ind + 1)) return true;vis[x][y] = false;return false;}
}
写法二
for循环
class Solution {int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};boolean[][] vis;public boolean exist(char[][] board, String word) {int m = board.length, n = board[0].length;vis = new boolean[m][n];for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){if(dfs(board, i, j, word, 0)) return true;}}return false;}boolean dfs(char[][] board, int x, int y, String word, int ind){int m = board.length, n = board[0].length;if(x < 0 || x >= m || y < 0 || y >= n || vis[x][y] || board[x][y] != word.charAt(ind)) return false;if(ind == word.length() - 1) return true;vis[x][y] = true;for(int i = 0; i < dir.length; i++){int nx = x + dir[i][0];int ny = y + dir[i][1];if(dfs(board, nx, ny, word, ind + 1)) return true;}vis[x][y] = false;return false;}
}之前写法
解法一和解法二 vis数组 写法不同。
解法一处理的是节点，即进入该节点后，更改当前节点的访问状态。
解法二处理的是边，即进入该节点前，更改即将进入节点的访问状态。
需要注意的是，两种写法不要混淆。
解法一
class Solution {int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};boolean[][] vis;public boolean exist(char[][] board, String word) {int m = board.length, n = board[0].length;vis = new boolean[m][n];for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){if(backtrack(board, i, j, word, 0)) return true;}}return false;}boolean backtrack(char[][] board, int x, int y, String word, int ind){int m = board.length, n = board[0].length;if(board[x][y] != word.charAt(ind)) return false;if(ind == word.length() - 1) return true;vis[x][y] = true;for(int i = 0; i < dir.length; i++){int nx = x + dir[i][0];int ny = y + dir[i][1];if(nx < 0 || nx >= m || ny < 0 || ny >= n || vis[nx][ny]) continue;if(backtrack(board, nx, ny, word, ind + 1)) return true;}vis[x][y] = false;return false;}
}
解法二
class Solution {int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};boolean[][] vis;public boolean exist(char[][] board, String word) {int m = board.length, n = board[0].length;vis = new boolean[m][n];for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){vis[i][j] = true;if(backtrack(board, i, j, word, 0)) return true;vis[i][j] = false;}}return false;}boolean backtrack(char[][] board, int x, int y, String word, int ind){int m = board.length, n = board[0].length;if(board[x][y] != word.charAt(ind)) return false;if(ind == word.length() - 1) return true;for(int k = 0; k < dir.length; k++){int nx = x + dir[k][0];int ny = y + dir[k][1];if(nx < 0 || nx >= m || ny < 0 || ny >= n || vis[nx][ny]) continue;vis[nx][ny] = true;if(backtrack(board, nx, ny, word, ind + 1)) return true;vis[nx][ny] = false;}return false;}
}
)
