724. Find Pivot Index

題目敘述

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 <= 10**4
• -1000 <= nums[i] <= 1000

Note: This question is the same as 1991: https://leetcode.com/problems/find-the-middle-index-in-array/

Hint 1:

We can precompute prefix sums P[i] = nums[0] + nums[1] + ... + nums[i-1]. Then for each index, the left sum is P[i], and the right sum is P[P.length - 1] - P[i] - nums[i].

題目翻譯

這題是要找出中間錨點的索引，中間錨點會拆分左右兩邊，其兩側的總和需要相等。如果沒有滿足條件的位置，就回傳 -1。如果有多個滿足條件的錨點，就回傳最左側的。

解法解析

此題與 1991 相同。 利用左右兩側相同的原則 sum = leftsum + rightsum + x，就會等於 sum = 2 * leftsum + x。所以先找出總和，然後再比對。

解法範例

Go

func pivotIndex(nums []int) int {
    var S int = 0
    for _, v := range nums {
        S += v
    }
    var leftsum int = 0
    for i, x := range nums {
        if leftsum == S-leftsum-x {
            return i
        }
        leftsum += x
    }
    return -1
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var pivotIndex = function (nums) {
    const S = nums.reduce((a, b) => a + b, 0);
    let leftsum = 0;
    for (let i = 0; i < nums.length; i++) {
        if (leftsum === S - leftsum - nums[i]) {
            return i;
        }
        leftsum += nums[i];
    }
    return -1;
};

Kotlin

class Solution {
    fun pivotIndex(nums: IntArray): Int {
        val sum = nums.sum()
        var leftSum = 0
        for ((i, num) in nums.withIndex()) {
            if (leftSum == (sum - leftSum - num)) {
                return i
            }
            leftSum += num
        }
        return -1
    }
}

PHP

class Solution
{

    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function pivotIndex($nums)
    {
        $sum = array_sum($nums);
        $leftSum = 0;
        foreach ($nums as $i => $num) {
            if ($leftSum == $sum - $leftSum - $num) {
                return $i;
            }
            $leftSum += $num;
        }
        return -1;
    }
}

Python

class Solution:
    def pivotIndex(self, nums: List[int]) -> int:
        S = sum(nums)
        leftsum = 0
        for i, x in enumerate(nums):
            if leftsum == (S - leftsum - x):
                return i
            leftsum += x
        return -1

Rust

Swift

class Solution {
    func pivotIndex(_ nums: [Int]) -> Int {
        let sum = nums.reduce(0, +)
        var leftSum = 0
        for (i, num) in nums.enumerated() {
            if leftSum == sum - leftSum - num {
                return i
            }
            leftSum += num
        }
        return -1
    }
}