Find Pivot Index

最新推荐文章于 2024-01-03 19:33:05 发布

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本文介绍了一个寻找数组中心索引的问题，即找到数组中一个特定的位置，使得该位置左侧所有元素之和等于右侧所有元素之和。文章提供了一种有效的解决方案，并通过一个C++示例展示了如何实现。

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POBLEM：

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].

SOLVE：

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

分析：一开始想当然以为数字全是整数，以为能缩短时间复杂度直接写了一种错误的算法...要考虑负面输入！