LeetCode 79. Word Search
原题链接在这里：https://leetcode.com/problems/word-search/
题目：
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 =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
题解：
类似N-Queens. dfs, 终止条件若能走到index == word.length(), 返回true.
若是做不到就是中间出现了越界或者不等，返回false.
用之前要把used 对应位置改为true, 用完后记得把used 对应位置改为false.
最后返回res.
Note: 1. Check condition order. First check index == word.length(), since there could be case that word found, should return true, but index is out of bound and return false.
2. Revert the boolean visited map before each return.
Time Complexity: O(m^2 * n^2). 对于board每一个点search都是一次dfs, dfs用了O(V+E), 矩阵表示时就是O(m*n). 并且board一共有m*n个点.
Space: O(m*n). 因为用了used存储。这里m=board.length, n = board[0].length.
AC Java:
1 public class Solution { 2 public boolean exist(char[][] board, String word) { 3 if(word==null || word.length()==0) 4 return true; 5 if(board==null || board.length==0 || board[0].length==0) 6 return false; 7 boolean[][] used = new boolean[board.length][board[0].length]; 8 for(int i=0;i<board.length;i++) 9 { 10 for(int j=0;j<board[0].length;j++) 11 { 12 if(search(board,word,0,i,j,used)) 13 return true; 14 } 15 } 16 return false; 17 } 18 private boolean search(char[][] board, String word, int index, int i, int j, boolean[][] used) 19 { 20 if(index == word.length()) 21 return true; 22 if(i<0 || j<0 || i>=board.length || j>=board[0].length || used[i][j] || board[i][j]!=word.charAt(index)) 23 return false; 24 used[i][j] = true; 25 boolean res = search(board,word,index+1,i-1,j,used) 26 || search(board,word,index+1,i+1,j,used) 27 || search(board,word,index+1,i,j-1,used) 28 || search(board,word,index+1,i,j+1,used); 29 used[i][j] = false; 30 return res; 31 } 32 }
