首页 > itarticle > leetcode笔记：37. sudoku solver

leetcode笔记：37. sudoku solver

admin 11月 14, 2020 0

#	Title	Difficulty	Topic
37	Sudoku Solver	Hard	Hash Table; Backtracking

Description

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.

Example:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

Backtracking

class  {
    public boolean exist(char[][] board, String word) {
        if(board == null || board.length == 0 ||
           board[0] == null || board[0].length == 0) return false;
        char[] words = word.toCharArray();
        int m = board.length, n = board[0].length;
        if(m * n < words.length) return false;
        
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(helper(board, i, j, words, 0)) return true;
            }
        }
        return false;
    }
    
    public boolean helper(char[][] board, int i, int j, char[] words, int start) {
        if(start == words.length) return true;
        if(i >= board.length || i < 0 ||
           j >= board[0].length || j < 0) return false;
        if(board[i][j] != words[start]) return false;
        
        boolean exist = false;
        board[i][j] += 128; 
        if(helper(board, i-1, j, words, start+1) || 
           helper(board, i+1, j, words, start+1) ||
           helper(board, i, j-1, words, start+1) ||
           helper(board, i, j+1, words, start+1)) exist = true;
        board[i][j] -= 128; // board[i][j] &= 127;
        return exist;
    }
}

时间复杂度为$$O(mn 4^L)$$，空间复杂度$$O(L)$$，$$L$$为word的长度。