| comments | difficulty | edit_url |
|---|---|---|
true |
中等 |
一只蚂蚁坐在由白色和黑色方格构成的无限网格上。开始时，网格全白，蚂蚁面向右侧。每行走一步，蚂蚁执行以下操作。
(1) 如果在白色方格上，则翻转方格的颜色，向右(顺时针)转 90 度，并向前移动一个单位。
(2) 如果在黑色方格上，则翻转方格的颜色，向左(逆时针方向)转 90 度，并向前移动一个单位。
编写程序来模拟蚂蚁执行的前 K 个动作，并返回最终的网格。
网格由数组表示，每个元素是一个字符串，代表网格中的一行，黑色方格由 'X' 表示，白色方格由 '_' 表示，蚂蚁所在的位置由 'L', 'U', 'R', 'D' 表示，分别表示蚂蚁 左、上、右、下 的朝向。只需要返回能够包含蚂蚁走过的所有方格的最小矩形。
示例 1:
输入: 0 输出: ["R"]
示例 2:
输入: 2 输出: [ "_X", "LX" ]
示例 3:
输入: 5 输出: [ "_U", "X_", "XX" ]
说明：
K <= 100000
我们使用哈希表
我们模拟蚂蚁的行走过程。如果蚂蚁所在的方格是白色的，那么蚂蚁向右转
模拟结束后，我们根据
时间复杂度
class Solution:
def printKMoves(self, K: int) -> List[str]:
x1 = y1 = x2 = y2 = 0
dirs = (0, 1, 0, -1, 0)
d = "RDLU"
x = y = 0
p = 0
black = set()
for _ in range(K):
if (x, y) in black:
black.remove((x, y))
p = (p + 3) % 4
else:
black.add((x, y))
p = (p + 1) % 4
x += dirs[p]
y += dirs[p + 1]
x1 = min(x1, x)
y1 = min(y1, y)
x2 = max(x2, x)
y2 = max(y2, y)
m, n = x2 - x1 + 1, y2 - y1 + 1
g = [["_"] * n for _ in range(m)]
for i, j in black:
g[i - x1][j - y1] = "X"
g[x - x1][y - y1] = d[p]
return ["".join(row) for row in g]class Solution {
public List<String> printKMoves(int K) {
int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
int[] dirs = {0, 1, 0, -1, 0};
String d = "RDLU";
int x = 0, y = 0, p = 0;
Set<List<Integer>> black = new HashSet<>();
while (K-- > 0) {
List<Integer> t = List.of(x, y);
if (black.add(t)) {
p = (p + 1) % 4;
} else {
black.remove(t);
p = (p + 3) % 4;
}
x += dirs[p];
y += dirs[p + 1];
x1 = Math.min(x1, x);
y1 = Math.min(y1, y);
x2 = Math.max(x2, x);
y2 = Math.max(y2, y);
}
int m = x2 - x1 + 1;
int n = y2 - y1 + 1;
char[][] g = new char[m][n];
for (char[] row : g) {
Arrays.fill(row, '_');
}
for (List<Integer> t : black) {
int i = t.get(0) - x1;
int j = t.get(1) - y1;
g[i][j] = 'X';
}
g[x - x1][y - y1] = d.charAt(p);
List<String> ans = new ArrayList<>();
for (char[] row : g) {
ans.add(String.valueOf(row));
}
return ans;
}
}class Solution {
public:
vector<string> printKMoves(int K) {
int x1 = 0, y1 = 0, x2 = 0, y2 = 0;
int dirs[5] = {0, 1, 0, -1, 0};
string d = "RDLU";
int x = 0, y = 0, p = 0;
set<pair<int, int>> black;
while (K--) {
auto t = make_pair(x, y);
if (black.count(t)) {
black.erase(t);
p = (p + 3) % 4;
} else {
black.insert(t);
p = (p + 1) % 4;
}
x += dirs[p];
y += dirs[p + 1];
x1 = min(x1, x);
y1 = min(y1, y);
x2 = max(x2, x);
y2 = max(y2, y);
}
int m = x2 - x1 + 1, n = y2 - y1 + 1;
vector<string> g(m, string(n, '_'));
for (auto& [i, j] : black) {
g[i - x1][j - y1] = 'X';
}
g[x - x1][y - y1] = d[p];
return g;
}
};func printKMoves(K int) []string {
var x1, y1, x2, y2, x, y, p int
dirs := [5]int{0, 1, 0, -1, 0}
d := "RDLU"
type pair struct{ x, y int }
black := map[pair]bool{}
for K > 0 {
t := pair{x, y}
if black[t] {
delete(black, t)
p = (p + 3) % 4
} else {
black[t] = true
p = (p + 1) % 4
}
x += dirs[p]
y += dirs[p+1]
x1 = min(x1, x)
y1 = min(y1, y)
x2 = max(x2, x)
y2 = max(y2, y)
K--
}
m, n := x2-x1+1, y2-y1+1
g := make([][]byte, m)
for i := range g {
g[i] = make([]byte, n)
for j := range g[i] {
g[i][j] = '_'
}
}
for t := range black {
i, j := t.x-x1, t.y-y1
g[i][j] = 'X'
}
g[x-x1][y-y1] = d[p]
ans := make([]string, m)
for i := range ans {
ans[i] = string(g[i])
}
return ans
}class Solution {
func printKMoves(_ K: Int) -> [String] {
var x1 = 0, y1 = 0, x2 = 0, y2 = 0
let dirs = [0, 1, 0, -1, 0]
let d = "RDLU"
var x = 0, y = 0, p = 0
var black = Set<[Int]>()
var K = K
while K > 0 {
let t = [x, y]
if black.insert(t).inserted {
p = (p + 1) % 4
} else {
black.remove(t)
p = (p + 3) % 4
}
x += dirs[p]
y += dirs[p + 1]
x1 = min(x1, x)
y1 = min(y1, y)
x2 = max(x2, x)
y2 = max(y2, y)
K -= 1
}
let m = x2 - x1 + 1
let n = y2 - y1 + 1
var g = Array(repeating: Array(repeating: "_", count: n), count: m)
for t in black {
let i = t[0] - x1
let j = t[1] - y1
g[i][j] = "X"
}
g[x - x1][y - y1] = String(d[d.index(d.startIndex, offsetBy: p)])
return g.map { $0.joined() }
}
}