【LeetCode】724. Find Pivot Index（C++）

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本文探讨了如何在给定的整数数组中找到“中心”索引，该索引左边所有元素的总和等于右边所有元素的总和。通过分析LeetCode上的经典问题，提供了一种有效的解决方案，并对比了不同算法的适用场景。

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地址：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:

All the strings in the input will only contain lowercase letters.

The length of nums will be in the range [0, 10000].
Each element nums[i] will be an integer in the range [-1000, 1000].

理解：

找到一个位置i，使得

nums[0]+···+nums[i-1]==nums[i+1]+···+nums[nums.size()-1]

实现：

最开始是用下面的方式实现的，但是这种只适用于数组非负的情况，双指针，一个从左开始加一个从右开始加。

class Solution {
public:
	int pivotIndex(vector<int>& nums) {
		int i = 0, j = nums.size() - 1;
		int sum1 = 0, sum2 = 0;
		while (i<j) {
			if (sum1<sum2) {
				sum1 += nums[i];
				++i;
			}
			else if (sum1>sum2) {
				sum2 += nums[j];
				--j;
			}
			else {
				sum1 += nums[i];
				++i;
				sum2 += nums[j];
				--j;
			}
		}
		if (i == j&&sum1 == sum2)
			return i;
		else
			return -1;
	}
};

当存在负值的时候就不能用了，比如nums = [-1, -1, -1, -1, -1, 0]。
实际上，设数组的和为sum，i左侧的和为leftSum，需要判断leftSum==sum-nums[i]-leftSum是否为真，即可。
实现如下：

class Solution {
public:
	int pivotIndex(vector<int>& nums) {
		int sum = 0, leftSum = 0;
		for (int i : nums)
			sum += i;
		for (int i = 0; i < nums.size(); ++i) {
			if (leftSum == sum - leftSum - nums[i])
				return i;
			leftSum += nums[i];
		}
		return -1;
	}
};