每天一道Rust-LeetCode(2019-11-21) ​

坚持每天一道题,刷题学习Rust.

题目描述 ​

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。 数字 1-9 在每一列只能出现一次。 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:

输入: [ ["5","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 输出: true 示例 2:

输入: [ ["8","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 输出: false 解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。 说明:

一个有效的数独（部分已被填充）不一定是可解的。 只需要根据以上规则，验证已经填入的数字是否有效即可。 给定数独序列只包含数字 1-9 和字符 '.' 。 给定数独永远是 9x9 形式的。

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/valid-sudoku 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解题思路 ​

思路: 看题目觉得需要三遍遍历,答案给出了一遍遍历,初看没看明白 原来是每行,每列都有一个独立的map来存储计数,每个小子宫格也是如此 这样确实是一次遍历,但是存储空间却浪费了更多, 个人感觉还不如逐行逐列判断. 但是如果是在动态的求数独的过程中,显然一遍遍历的方式更合适.

解题过程 ​

use std::collections::HashMap;

struct Solution;

impl Solution {
    pub fn is_valid_sudoku(board: Vec<Vec<char>>) -> bool {
        let mut rows = Vec::new();
        let mut cols = Vec::new(); //行列
        let mut boxes = Vec::new();
        for _ in 0..9 {
            rows.push(HashMap::new());
            cols.push(HashMap::new());
            boxes.push(HashMap::new());
        }
        for i in 0..9 {
            for j in 0..9 {
                let c = board[i][j];
                //println!("test={}",c);
                if c == '.' {
                    continue;
                }
                let bi = (i / 3) * 3 + j / 3;
                if rows[i].contains_key(&c)
                    || cols[j].contains_key(&c)
                    || boxes[bi].contains_key(&c)
                {
                    return false;
                }
                rows[i].insert(c, 1);
                cols[j].insert(c, 1);
                boxes[bi].insert(c, 1);
            }
        }
        true
    }
}
 

一点感悟 ​

map是一个好东西啊,怪不得go把他放到标准库里,并且使用语法糖

其他 ​

欢迎关注我的github,本项目文章所有代码都可以找到.