标签：style c class blog code java
DFS算法，非常容易TLE，需要一个boolean矩阵来记录是否访问过某个节点。
写DFS主要就是两个方法：用recursion或者Stack, 用recursion会带来time和memory的cost增加，而且因为要用矩阵做argument, 所以非常担心TLE或者MLE的问题。但是用recursion就比较容易使用boolean矩阵，可以比较容易地对某个节点设置visited或者unvisited。而用stack来写呢，cost会小，但是很难用boolean矩阵来设置之前访问过的某个节点为unvisited。所以思来想去，还是用recursion来写。
1 public class Solution { 2 private int m; 3 private int n; 4 public boolean exist(char[][] board, String word) { 5 m = board.length; 6 n = board[0].length; 7 boolean[][] visited = new boolean[m][n]; 8 for (int i = 0; i < m; i++) { 9 for (int j = 0; j < n; j++) { 10 if (board[i][j] == word.charAt(0)) { 11 if (dfs(board, i, j, visited, word, 0)) return true; 12 } 13 } 14 } 15 return false; 16 } 17 18 public boolean dfs(char[][] board, int x, int y, boolean[][] visited, String word, int k) { 19 if (k == word.length()) return true; 20 if (x < 0 || y < 0 || x >= m || y >= n) return false; 21 if (visited[x][y] == true) return false; 22 if (board[x][y] != word.charAt(k)) return false; 23 visited[x][y] = true; 24 boolean res = dfs(board, x, y-1, visited, word, k+1) 25 || dfs(board, x, y+1, visited, word, k+1) 26 || dfs(board, x-1, y, visited, word, k+1) 27 || dfs(board, x+1, y, visited, word, k+1); 28 visited[x][y] = false; 29 return res; 30 } 31 }
下面这个code TLE了，但是跟上面那个都没什么区别，不知道为什么
public class Solution { private int m; private int n; public boolean exist(char[][] board, String word) { m = board.length; n = board[0].length; boolean[][] visited = new boolean[m][n]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (board[i][j] == word.charAt(0)) { if (dfs(board, i, j, visited, word, 0)) return true; } } } return false; } public boolean dfs(char[][] board, int x, int y, boolean[][] visited, String word, int k) { boolean bool1 = false, bool2 = false, bool3 = false, bool4 = false; if (k == word.length()) return true; if (x < 0 || y < 0 || x >= m || y >= n) return false; if (visited[x][y] == true) return false; if (board[x][y] != word.charAt(k)) return false; visited[x][y] = true; bool1 = dfs(board, x, y-1, visited, word, k+1); //left bool2 = dfs(board, x, y+1, visited, word, k+1); //right bool3 = dfs(board, x-1, y, visited, word, k+1); //up bool4 = dfs(board, x+1, y, visited, word, k+1); //down visited[x][y] = false; return bool1 || bool2 || bool3 || bool4; } }
再贴一个用stack写的DFS，但是想不出来怎么设置之前访问过的一个节点为unvisited
1 public class Solution { 2 public boolean exist(char[][] board, String word) { 3 int m = board.length; 4 int n = board[0].length; 5 Stack<Integer> toVisit = new Stack<Integer>(); 6 boolean[][] visited = new boolean[m][n]; 7 if (word == "") return true; 8 if (m == 0 && n == 0 && word != "") return false; 9 for (int i = 0; i < m; i++) { 10 for (int j = 0; j < n; j++) { 11 if (board[i][j] == word.charAt(0)) { 12 toVisit.push(i * n + j); 13 } 14 } 15 } 16 17 int k = 0; 18 while (!toVisit.empty()) { 19 int root = toVisit.pop(); 20 int x = root / n; 21 int y = root % n; 22 visited[x][y] = true; 23 if (k == word.length() - 1) return true; 24 if (y > 0 && visited[x][y-1] == false && board[x][y-1] == word.charAt(k+1)) { //left 25 toVisit.push(x * n + y - 1); 26 k++; 27 } 28 if (y < n && visited[x][y+1] == false && board[x][y+1] == word.charAt(k+1)) { //right 29 toVisit.push(x * n + y + 1); 30 k++; 31 } 32 if (x > 0 && visited[x-1][y] == false && board[x-1][y] == word.charAt(k+1)) { //up 33 toVisit.push((x - 1) * n + y); 34 k++; 35 } 36 if (x < m && visited[x+1][y] == false && board[x+1][y] == word.charAt(k+1)) { //down 37 toVisit.push((x + 1) * n + y); 38 k++; 39 } 40 } 41 return false; 42 } 43 }
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标签：style c class blog code java
原文地址：http://www.cnblogs.com/EdwardLiu/p/3746765.html
