原题链接在这里：https://leetcode.com/problems/find-pivot-index/
题目：
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
numswill be in the range[0, 10000]. - Each element
nums[i]will be an integer in the range[-1000, 1000].
题解：
From left to right, accumlate the sum and update orinigal nums array.
For index = 0 and n - 1 position, we could treat sum of left of index 0 is 0. sum of right of index n - 1 is 0.
Then from left to right, when sum[i - 1] == sum[n - 1] - sum[i], this is the pivot index.
Time Complexity: O(n). n = nums.length.
Space: O(1).
AC Java:
1 class Solution { 2 public int pivotIndex(int[] nums) { 3 if(nums == null || nums.length < 3){ 4 return -1; 5 } 6 7 int n = nums.length; 8 9 for(int i = 1; i<n; i++){ 10 nums[i] += nums[i - 1]; 11 } 12 13 for(int i = 0; i<n; i++){ 14 int left = i == 0 ? 0 : nums[i - 1]; 15 if(left == nums[n - 1] - nums[i]){ 16 return i; 17 } 18 } 19 20 return -1; 21 } 22 }