题目：原题链接（中等）
标签：数组
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( K ) | O ( K ) | 736ms (62.73%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一（情景模拟）：
class Solution:
def printKMoves(self, K: int) -> List[str]:
if K == 0:
return ["R"]
# x0, y0, x1, y1 = 0, 0, 0, 0 # 四边坐标
black = set()
x, y = 0, 0
last = 0
stat = 0 # 0=R,1=D,2=L,3=U
while K >= 0:
# 计算新的坐标
if stat == 0:
x += 1
elif stat == 1:
y -= 1
elif stat == 2:
x -= 1
else:
y += 1
# 更新新的边缘位置
# x0, y0, x1, y1 = min(x, x0), min(y, y0), max(x, x1), max(y, y1)
last = stat
# 计算方向并调整地图颜色
if (x, y) in black:
stat = (stat - 1) % 4
black.remove((x, y))
else:
stat = (stat + 1) % 4
black.add((x, y))
K -= 1
# print(x, y, ["R", "D", "L", "U"][stat], "(", x0, y0, x1, y1, ")")
# 计算地图边缘位置
x0, y0, x1, y1 = float("inf"), float("inf"), float("-inf"), float("-inf") # 四边坐标
for i, j in black:
x0, y0, x1, y1 = min(i, x0), min(j, y0), max(i, x1), max(j, y1)
# 生成地图表
ans = []
for j in range(y1, y0 - 1, -1):
row = []
for i in range(x0, x1 + 1):
if i == x and j == y:
row.append(["R", "D", "L", "U"][last])
elif (i, j) in black:
row.append("X")
else:
row.append("_")
ans.append(row)
# 生成最终结果
return ["".join(row) for row in ans]
