Find Pivot Index
Given an array of integers nums, calculate the pivot index of this array.
The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index's right.
If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.
Return the leftmost pivot index. If no such index exists, return -1.
Example 1:
Input: nums = [1,7,3,6,5,6]
Output: 3
Explanation:
The pivot index is 3.
Left sum = nums[0] + nums[1] + nums[2] = 1 + 7 + 3 = 11
Right sum = nums[4] + nums[5] = 5 + 6 = 11
Example 2:
Input: nums = [1,2,3]
Output: -1
Explanation:
There is no index that satisfies the conditions in the problem statement.
Example 3:
Input: nums = [2,1,-1]
Output: 0
Explanation:
The pivot index is 0.
Left sum = 0 (no elements to the left of index 0)
Right sum = nums[1] + nums[2] = 1 + -1 = 0
Constraints:
1 <= nums.length <= 104
-1000 <= nums[i] <= 1000
思路一：先对整个数组求和，然后从头开始遍历数组，对比前后是否相等
public int pivotIndex(int[] nums) {
int sum = Arrays.stream(nums).sum();
int sumFromBegin = 0;
for (int i = 0; i < nums.length; i++) {
sum -= nums[i];
if (i > 0) {
sumFromBegin += nums[i - 1];
}
if (sumFromBegin == sum) {
return i;
}
}
return -1;
}
思路二：看了一下官解，思路更巧妙，左侧和*2+nums[i]=总和，不用求右侧和