Word Search

Change-Id: I38e4be18de4b4725b0ea836653f56a65a6f5f99a

Author: applewjg

WordSearch.java

@@ -0,0 +1,55 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 20, 2014
+ Problem:    Word Search
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/word-search/
+ Notes:
+ Given a 2D board and a word, find if the word exists in the grid.
+ The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are 
+ those horizontally or vertically neighboring. The same letter cell may not be used more than once.
+ For example,
+ Given board =
+ [
+  ["ABCE"],
+  ["SFCS"],
+  ["ADEE"]
+ ]
+ word = "ABCCED", -> returns true,
+ word = "SEE", -> returns true,
+ word = "ABCB", -> returns false.
+
+ Solution: DFS.
+ */
+public class Solution {
+    public boolean exist(char[][] board, String word) {
+        int m = board.length;
+        if (m == 0) return false;
+        int n = board[0].length;
+        if (n == 0) return false;
+        if (word.length() == 0) return true;
+        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) && existRe(board, i, j, word, 0, visited)) {
+                    return true;
+                }
+            }
+        }
+        return false;
+    }
+    public boolean existRe(char[][] board, int i, int j, String word, int cur, boolean[][] visited) {
+        if (cur == word.length()) return true;
+        int m = board.length;
+        int n = board[0].length;
+        if (i < 0 || i >= m || j < 0 || j >= n) return false;
+        if (visited[i][j] == true || (board[i][j] != word.charAt(cur))) return false;
+        visited[i][j] = true;
+        if (existRe(board, i+1, j, word, cur+1,visited)) return true;
+        if (existRe(board, i-1, j, word, cur+1,visited)) return true;
+        if (existRe(board, i, j+1, word, cur+1,visited)) return true;
+        if (existRe(board, i, j-1, word, cur+1,visited)) return true;
+        visited[i][j] = false;
+        return false;
+    }
+}