题目：原题链接（中等）
标签：广度优先搜索
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( X × Y ) O(X×Y) O(X×Y) | O ( X × Y ) O(X×Y) O(X×Y) | 6356ms (16.16%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def minKnightMoves(self, x: int, y: int) -> int:
ans = 0
visited = {(0, 0)}
queue = collections.deque([(0, 0)])
while queue:
for _ in range(len(queue)):
x1, y1 = queue.popleft()
if (x1, y1) == (x, y):
return ans
for x2, y2 in [(x1 + 2, y1 - 1), (x1 + 2, y1 + 1), (x1 + 1, y1 + 2), (x1 + 1, y1 - 2), (x1 - 1, y1 + 2),
(x1 - 1, y1 - 2), (x1 - 2, y1 - 1), (x1 - 2, y1 + 1)]:
if (x2, y2) not in visited:
visited.add((x2, y2))
queue.append((x2, y2))
ans += 1
