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 ( l o g N ) O(logN) O(logN) | O ( 1 ) O(1) O(1) | 136ms (69.15%) |
Ans 2 (Python) | O ( l o g N ) O(logN) O(logN) | O ( 1 ) O(1) O(1) | 136ms (69.15%) |
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(二分查找):
def nextGreatestLetter(self, letters: List[str], target: str) -> str:
if ord(letters[-1]) <= ord(target):
return letters[0]
left = 0
right = len(letters) - 1
while left < right:
mid = (left + right) // 2
if ord(letters[mid]) <= ord(target):
left = mid + 1
else:
right = mid
return letters[left]
解法二(二分查找):
def nextGreatestLetter(self, letters: List[str], target: str) -> str:
left = 0
right = len(letters) - 1
while left <= right:
mid = (left + right) // 2
if ord(letters[mid]) <= ord(target):
left = mid + 1
else:
right = mid - 1
if left == len(letters):
return letters[0]
else:
return letters[left]