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find-pivot-index.py
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find-pivot-index.py
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# Time: O(n)
# Space: O(1)
# Given an array of integers nums, write a method that returns the "pivot" index of this array.
#
# We define the pivot index as the index where the sum of the numbers to
# the left of the index is equal to the sum of the numbers to the right of the index.
#
# If no such index exists, we should return -1. If there are multiple pivot indexes,
# you should return the left-most pivot index.
#
# Example 1:
# Input:
# nums = [1, 7, 3, 6, 5, 6]
# Output: 3
# Explanation:
# The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
# Also, 3 is the first index where this occurs.
#
# Example 2:
# Input:
# nums = [1, 2, 3]
# Output: -1
# Explanation:
# There is no index that satisfies the conditions in the problem statement.
#
# Note:
# - The length of nums will be in the range [0, 10000].
# - Each element nums[i] will be an integer in the range [-1000, 1000].
class Solution(object):
def pivotIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
total = sum(nums)
left_sum = 0
for i, num in enumerate(nums):
if left_sum == (total-left_sum-num):
return i
left_sum += num
return -1