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(logn) | O(1) | 40ms (61.99%) |
Ans 2 (Python) | O(logn) | O(1) | 36ms (83.47%) |
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(二分法查找):
def firstBadVersion(self, n):
left, right = 0, n
while (right - left) > 1:
binery = left + (right - left) // 2
binery_ans = isBadVersion(binery)
if binery_ans:
right = binery
else:
left = binery
return right
解法二(另一种二分查找):
def firstBadVersion(self, n):
left, right = 0, n
while right > left:
binery = left + (right - left) // 2
binery_ans = isBadVersion(binery)
if binery_ans:
right = binery
else:
left = binery + 1
return left