2014雅虎校园招聘：二维字符数组查找Yahoo！（软件开发岗）
要求输入的单词的字符之间在数组里面是邻接关系
例如 ：输入二维字符矩阵如下：
A H E L
K M O L
H B W P
查找HELLOW明显可以找到，对应的坐标为(0,1)——>(0,2)——>(0,3)——>(1,3)——>(1,2)——>(2,2)，相当于一笔画成HELLOW！
本题采用枚举法实现。动态查找每一位的邻接是否满足条件，适当的剪枝是必要的！
JAVA源码如下(没有对输入数据进行合法性的检查！)：
package com;public class SearchWord {public static void main(String[] args) {String[] a = { "hello2rt", "orow -=!","xldpoiyx","uu!yuuzz" };System.out.println(isWord(a, "hello world!"));}public static boolean isWord(String[] matrix, String word) {boolean ok = false;for (int i = 0; i < matrix.length; i++) {for (int j = 0; j < matrix[i].length(); j++) {ok = dfs(0, matrix, word, i, j);if (ok) break;}if (ok) break;}return ok;}public static boolean dfs(int index, String[] matrix, String word, int i,int j) {if (index >= word.length())return false;if (!(i >= 0 && i < matrix.length && j >= 0 && j < matrix[i].length()))return false;char cur = word.charAt(index);if (cur == matrix[i].charAt(j) && (index + 1) == word.length())return true;if (cur == matrix[i].charAt(j)) {if (dfs(index + 1, matrix, word, i - 1, j))return true;if (dfs(index + 1, matrix, word, i + 1, j))return true;if (dfs(index + 1, matrix, word, i, j - 1))return true;if (dfs(index + 1, matrix, word, i, j + 1))return true;}return false;}}