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) |
|
| 48ms (37.23%) |
Ans 2 (Python) |
|
| 32ms (98.85%) |
Ans 3 (Python) |
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 minCostToMoveChips(self, chips: List[int]) -> int:
a = 0
b = 0
for chip in chips:
if chip % 2 == 0:
a += 1
else:
b += 1
return min(a, b)
解法二(优化解法一):
def minCostToMoveChips(self, chips: List[int]) -> int:
a = 0
for chip in chips:
a += (chip % 2 == 0)
return min(a, len(chips) - a)