题目:原题链接(简单)
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | O ( N 2 ) O(N^2) O(N2) | O ( N ) O(N) O(N) | 48ms (47.30%) |
Ans 2 (Python) | O ( 1 ) O(1) O(1) | O ( 1 ) O(1) O(1) | 28ms (99.56%) |
Ans 3 (Python) | O ( 1 ) O(1) O(1) | O ( 1 ) O(1) O(1) | 36ms (91.69%) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(情景模拟):
def divisorGame(self, N: int) -> bool:
situations = {
1: False
}
for i in range(2, N + 1):
chooses = {i - 1}
for k in (2, i // 2 + 1):
if i % k == 0 and 0 < i - k < i:
chooses.add(i - k)
situations[i] = not all(situations[choose] for choose in chooses)
return situations[N]
解法二:
def divisorGame(self, N: int) -> bool:
return N % 2 == 0
解法三(位运算):
def divisorGame(self, N: int) -> bool:
return not N & 1