LeetCode_79 Word Search
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Link to original problem: 这里写链接内容
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.
Related problem:
212 Word Search II: 这里写链接内容
关联问题212中提到了使用Trie数据结构。
假设board的大小为m*n,给定word长度为k的话,那么使用背包搜索的时间复杂度应为O(m*n*4^k),与以往的n!的算法的复杂度略有不同。以下是代码:
public class Solution { public boolean exist(char[][] board, String word) { if(word == null || word.length() == 0) return true; if(board == null || board.length == 0 || board[0].length == 0) return false; boolean[][] visited = new boolean[board.length][board[0].length]; char[] str = word.toCharArray(); for(int ii = 0; ii < board.length; ii++){ for(int jj = 0; jj < board[0].length; jj++){ if(helper(board, visited, str, 0, ii, jj)) return true; } } return false; } private boolean helper(char[][] board, boolean[][] visited, char[] str, int index, int i, int j){ if(index == str.length) return true; if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || visited[i][j] || board[i][j] != str[index]) return false; for(int kk = 0; kk < 4; kk++){ visited[i][j] = true; if(helper(board, visited, str, index+1, i+(kk%2)*(kk-2), j+((kk+1)%2)*(kk-1))) return true; visited[i][j] = false; } return false; }}
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