标签：cti rap 存在 common col alt continue nta als
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。
1-9 在每一行只能出现一次。1-9 在每一列只能出现一次。1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。
下面是Rust测试用的数组，可以复制过去。
let vec: Vec<Vec<char>> = vec![
vec![‘5‘, ‘3‘, ‘.‘, ‘.‘, ‘7‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘],
vec![‘6‘, ‘.‘, ‘.‘, ‘1‘, ‘9‘, ‘5‘, ‘.‘, ‘.‘, ‘.‘],
vec![‘.‘, ‘9‘, ‘8‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘6‘, ‘.‘],
vec![‘8‘, ‘.‘, ‘.‘, ‘.‘, ‘6‘, ‘.‘, ‘.‘, ‘.‘, ‘3‘],
vec![‘4‘, ‘.‘, ‘.‘, ‘8‘, ‘.‘, ‘3‘, ‘.‘, ‘.‘, ‘1‘],
vec![‘7‘, ‘.‘, ‘.‘, ‘.‘, ‘2‘, ‘.‘, ‘.‘, ‘.‘, ‘6‘],
vec![‘.‘, ‘6‘, ‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘2‘, ‘8‘, ‘.‘],
vec![‘.‘, ‘.‘, ‘.‘, ‘4‘, ‘1‘, ‘9‘, ‘.‘, ‘.‘, ‘5‘],
vec![‘.‘, ‘.‘, ‘.‘, ‘.‘, ‘8‘, ‘.‘, ‘.‘, ‘7‘, ‘9‘],
];
题解： 这个题就是一堆数组查重，我的思路是用变量去表示每个元素，过滤掉 ‘.‘,然后放入map，判断是否存在，下面是代码 4ms；
use std::collections::HashMap;
let mut j = 0; // 行
let mut k = 0; // 列
for i in 0..9 {
let mut map: HashMap<char, char> = HashMap::new();
let mut map2: HashMap<char, char> = HashMap::new();
let mut map3: HashMap<char, char> = HashMap::new();
for j in 0..9 {
if board[i][j] != ‘.‘ {
if map.contains_key(&board[i][j]) {
return false;
}
map.insert(board[i][j], ‘0‘);
}
if board[j][i] != ‘.‘ {
if map2.contains_key(&board[j][i]) {
return false;
}
map2.insert(board[j][i], ‘0‘);
}
}
if k == 9 {
k = 0;
j = j + 3;
}
let mut x = j;
let mut y = k;
for _ in 0..9 {
if board[x][y] != ‘.‘ {
if map3.contains_key(&board[x][y]) {
return false;
}
map3.insert(board[x][y], ‘0‘);
}
if y == 2 {
x = x + 1;
y = 0;
continue;
}
if y == 5 {
x = x + 1;
y = 3;
continue;
}
if y == 8 {
x = x + 1;
y = 6;
continue;
}
y = y + 1
}
k = k + 3;
}
return true;
大佬的解法，看着更赏心悦目 0ms，原理都差不多，只是放在了数组，这种查重时候 0..10 的数组
let mut row = [[0u8; 10]; 9];
let mut col = [[0u8; 10]; 9];
let mut my_box = [[0u8; 10]; 9];
for i in 0..9 {
for j in 0..9 {
if board[i][j] == ‘.‘ {
continue;
};
let cur_number = board[i][j].to_string().parse::<usize>().unwrap();
print!("{} ", cur_number);
if row[i][cur_number] != 0 {
return false;
}
if col[j][cur_number] != 0 {
return false;
}
if my_box[j / 3 + (i / 3) * 3][cur_number] != 0 {
return false;
}
row[i][cur_number] = 1;
col[j][cur_number] = 1;
my_box[j / 3 + (i / 3) * 3][cur_number] = 1;
}
}
return true;
标签：cti rap 存在 common col alt continue nta als
原文地址：https://www.cnblogs.com/Addoil/p/13396035.html
