Word Search

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.

说实话这题做的挺瞎的，题目挺简单，DFS呗，用递归做可能根本用不着写这么麻烦，就是怕递归超时才转成非递归，结果写的巨麻烦，直接贴代码吧，想看的看一看就好了。

 1 public boolean exist(char[][] board, String word) {
 2         int grid[][] = new int[board.length][board[0].length];
 3         for (int i = 0; i < board.length; i++) {
 4             for (int j = 0; j < board[0].length; j++) {
 5                 if (word.charAt(0) == board[i][j]) {
 6                     grid[i][j] = 1;
 7                     Stack<Info> stack = new Stack<Info>();
 8                     stack.push(new Info(i, j, 0, 0, 0));
 9                     while (!stack.empty()) {
10                         Info info = stack.pop();
11                         if (info.pos == word.length() - 1) {
12                             return true;
13                         }
14                         if(info.time==4){
15                             grid[info.i][info.j]=0;
16                             continue;
17                         }
18                         if (info.time<1&&info.direct != 1
19                                 && info.j > 0
20                                 && grid[info.i][info.j - 1] == 0
21                                 && word.charAt(info.pos + 1) == board[info.i][info.j - 1]) {
22                             info.time=1;
23                             stack.push(info);
24                             stack.push(new Info(info.i, info.j - 1, 2,
25                                     info.pos + 1,0));
26                             grid[info.i][info.j - 1] = 1;
27                         } else if (info.time<2&&info.direct != 2
28                                 && info.j < board[0].length - 1
29                                 && grid[info.i][info.j + 1] == 0
30                                 && word.charAt(info.pos + 1) == board[info.i][info.j + 1]) {
31                             info.time=2;
32                             stack.push(info);
33                             stack.push(new Info(info.i, info.j + 1, 1,
34                                     info.pos + 1,0));
35                             grid[info.i][info.j + 1] = 1;
36                         } else if (info.time<3&&info.direct != 3
37                                 && info.i > 0
38                                 && grid[info.i - 1][info.j] == 0
39                                 && word.charAt(info.pos + 1) == board[info.i - 1][info.j]) {
40                             info.time=3;
41                             stack.push(info);
42                             stack.push(new Info(info.i - 1, info.j, 4,
43                                     info.pos + 1,0));
44                             grid[info.i - 1][info.j] = 1;
45                         } else if (info.time<4&&info.direct != 4
46                                 && info.i < board.length - 1
47                                 && grid[info.i + 1][info.j] == 0
48                                 && word.charAt(info.pos + 1) == board[info.i + 1][info.j]) {
49                             info.time=4;
50                             stack.push(info);
51                             stack.push(new Info(info.i + 1, info.j, 3,
52                                     info.pos + 1,0));
53                             grid[info.i + 1][info.j] = 1;
54                         } else {
55                             info.time=4;
56                             stack.push(info);
57                         }
58                     }
59                     grid[i][j] = 0;
60                 }
61             }
62         }
63         return false;
64     }

View Code

Info是用于记录状态信息的结构，i,j表示横纵坐标，pos表示与word的第几个相匹配，direct表示上一个匹配在这个匹配的哪个方向，time是当前点已经判断了那些下一步的方向。

 1 class Info {
 2     int i;
 3     int j;
 4     int direct;
 5     int pos;
 6     int time;
 7 
 8     Info() {
 9         i = 0;
10         j = 0;
11         direct = 0;
12         pos = 0;
13         time = 0;
14     }
15 
16     Info(int i, int j, int dierct, int pos, int time) {
17         this.i = i;
18         this.j = j;
19         this.direct = dierct;
20         this.pos = pos;
21         this.time = time;
22     }
23 
24     public String toString() {
25         return "" + i + " " + j + " " + direct + " " + pos+" "+time;
26     }
27 }

View Code