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  • Java for LeetCode 036 Valid Sudoku

    Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

    The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

    解题思路:

    传说中的数独(九宫格)问题,老实遍历三个规则即可:

    JAVA实现:

    	static public boolean isValidSudoku(char[][] board) {
    		for (int i = 0; i < board.length; i++) {
    			HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
    			for (int j = 0; j < board[0].length; j++) {
    				if (board[i][j] != '.') {
    					if (hashmap.containsKey(board[i][j]))
    						return false;
    					hashmap.put(board[i][j], 1);
    				}
    			}
    		}
    		for (int j = 0; j < board[0].length; j++) {
    			HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
    			for (int i = 0; i < board.length; i++) {
    				if (board[i][j] != '.') {
    					if (hashmap.containsKey(board[i][j]))
    						return false;
    					hashmap.put(board[i][j], 1);
    				}
    			}
    		}
    	    for (int i = 0; i < board.length; i += 3){
    	       for (int j = 0; j < board[0].length; j += 3){
    	           HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>();
    				for (int k = 0; k < 9; k++) {
    					if (board[i + k / 3][j + k % 3] != '.') {
    						if (hashmap.containsKey(board[i + k / 3][j + k % 3]))
    							return false;
    						hashmap.put(board[i + k / 3][j + k % 3], 1);
    					}
    				}
    			}
    		}
    		return true;
    	}
    
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  • 原文地址:https://www.cnblogs.com/tonyluis/p/4486600.html
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