原題鏈接在這里：https://leetcode.com/problems/word-search/

題目：

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.

題解：

類似N-Queens. dfs, 終止條件若能走到index == word.length(), 返回true.

若是做不到就是中間出現了越界或者不等，返回false.

用之前要把used 對應位置改為true, 用完后記得把used 對應位置改為false.?

最后返回res.

Time Complexity: O(m^2 * n^2). ?對于board每一個點search都是一次dfs, dfs用了O(V+E), 矩陣表示時就是O(m*n). 并且board一共有m*n個點.

Space: O(m*n). 因為用了used存儲。這里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 }

?跟上Word Search II

轉載于:https://www.cnblogs.com/Dylan-Java-NYC/p/4944270.html

總結

以上是生活随笔為你收集整理的LeetCode 79. Word Search的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。