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 ( 1 ) O(1) O(1) | 9104ms (11.77%) |
Ans 2 (Python) | O ( N ) O(N) O(N) | O ( 1 ) O(1) O(1) | 56ms (95.05%) |
Ans 3 (Python) | O ( N ) O(N) O(N) | O ( 1 ) O(1) O(1) | 64ms (86.25%) |
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（暴力解法）：
def pivotIndex(self, nums: List[int]) -> int:
for i in range(len(nums)):
if sum(nums[:i]) == sum(nums[i + 1:]):
return i
else:
return -1解法二（插值法）：
def pivotIndex(self, nums: List[int]) -> int:
differ = sum(nums)
for i in range(len(nums)):
if i > 0:
differ -= nums[i] + nums[i - 1]
else:
differ -= nums[i]
if differ == 0:
return i
else:
return -1解法三：
def pivotIndex(self, nums: List[int]) -> int:
total = sum(nums)
left = 0
for i in range(len(nums)):
if left == (total - left - nums[i]):
return i
left += nums[i]
else:
return -1
