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 nums will be in the range [0, 10000].
Each element nums[i] will be an integer in the range [-1000, 1000].
本题题意很简单，直接设置一个左右数组来求解左边元素和右边元素之和，然后遍历即可
代码如下：
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>
using namespace std;
class Solution
{
public:
int pivotIndex(vector<int>& nums)
{
int n = nums.size();
vector<int> left(n, 0);
for (int i = 1; i < n; i++)
left[i] = left[i - 1] + nums[i - 1];
vector<int> right(n, 0);
for (int i = n - 2; i >= 0; i--)
right[i] = right[i + 1] + nums[i + 1];
for (int i = 0; i < n; i++)
{
if (left[i] == right[i])
return i;
}
return -1;
}
};
