How to find the central index of an array by leetcode
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I. the content of the topic
Given an array nums of integer type, write a method that returns the array's "central index".
We define the central index of the array as follows: the sum of all elements on the left side of the central index of the array is equal to the sum of all elements on the right.
If the array does not have a central index, then we should return-1. If the array has more than one central index, then we should return the one closest to the left.
Example 1:
Enter:
Nums = [1,7,3,6,5,6]
Output: 3
Explanation:
The sum of the left number of index 3 (nums [3] = 6) (1 + 7 + 3 = 11) is equal to the sum of the right number (5 + 6 = 11).
At the same time, 3 is the first central index that meets the requirements.
Example 2:
Enter:
Nums = [1,2,3]
Output:-1
Explanation:
There is no central index in the array that meets this condition.
Description:
The length range of nums is [0, 10000].
Any nums [I] will be an integer in the range of [- 1000, 1000].
Second, the way to solve the problem
First calculate the sum of all elements in nums, and then calculate the left sum from left to right, if:
Left and * 2 + current element = all elements and
Returns the current element index, otherwise-1 is returned.
Code class Solution: def pivotIndex (self Nums: list)-> int: whole_sum = sum (nums) left_sum = 0 for i in range (len (nums)): if left_sum * 2 + nums [I] = whole_sum: return i else: left_sum + = nums [I] return-1if _ name__ = ='_ main__': S = Solution () nums1 = [1 7, 3, 6, 5, 6] ans1 = s.pivotIndex (nums1) print (ans1) nums2 = [1,2,3] ans2 = s.pivotIndex (nums2) print (ans2) nums3 = [- 1,-1,-1,-1,-1] ans3 = s.pivotIndex (nums3) print (ans3) nums4 = [- 1,-1,-1,-1,-1 0] ans4 = s.pivotIndex (nums4) print (ans4) these are all the contents of the article "how leetcode finds the central index of an array" Thank you for reading! I believe we all have a certain understanding, hope to share the content to help you, if you want to learn more knowledge, welcome to follow the industry information channel!
