https://leetcode.com/problems/find-pivot-index/description/
The length of
Each element
public int pivotIndex(int[] nums) {
int sum = 0, leftsum = 0;
for (int x: nums) sum += x;
for (int i = 0; i < nums.length; ++i) {
if (leftsum == sum - leftsum - nums[i]) return i;
leftsum += nums[i];
}
return -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:
nums will be in the range [0, 10000].nums[i] will be an integer in the range [-1000, 1000].int sum = 0, leftsum = 0;
for (int x: nums) sum += x;
for (int i = 0; i < nums.length; ++i) {
if (leftsum == sum - leftsum - nums[i]) return i;
leftsum += nums[i];
}
return -1;
}
public int pivotIndex(int[] nums) {
if (nums == null || nums.length == 0)
return -1;
long sum = Arrays.stream(nums).sum();
long prefixSum = 0;
for (int i = 0; i < nums.length; i++) {
if (sum - prefixSum - nums[i] == prefixSum) {
return i;
}
prefixSum += nums[i];
}
return -1;
}
