724. Find Pivot Index

最新推荐文章于 2024-01-03 19:33:05 发布

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最新推荐文章于 2024-01-03 19:33:05 发布

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原文链接：http://www.cnblogs.com/liuxinzhi/p/8137733.html

本文介绍了一种寻找数组中“中心”索引的方法。中心索引定义为数组左侧元素之和等于右侧元素之和的索引。文章提供了两种解决方案，并通过示例详细解释了如何实现。

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　　　　这道题为简单题

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

　　思路：

　　　　这个题比较简单，就是遍历整个数组就行了，但是我的代码太复杂，在网上找到了一个更简单的做法。

　　代码：

class Solution(object):
    def pivotIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) <= 1:
            return -1
        left = 0
        right = sum(nums) - nums[0]
        for i in range(len(nums)):
            if i == 0:
                if left == right: return 0
            elif i == len(nums) - 1:
                if sum(nums) - nums[len(nums)-1] == 0: return len(nums) - 1
            else:
                left += nums[i-1]
                right -= nums[i]
                if left == right:
                    return i
        return -1

　　　　大神思路和代码：

Input: [1, 7, 3, 6, 5, 6]

    index: 0, num: 1, left: 0, right: 27
    index: 1, num: 7, left: 1, right: 20
    index: 2, num: 3, left: 8, right: 17
    index: 3, num: 6, left: 11, right: 11 <-- Found!!!


class Solution(object):
    def pivotIndex(self, nums):
        # Time: O(n)
        # Space: O(1)
        left, right = 0, sum(nums)
        for index, num in enumerate(nums):
            right -= num
            if left == right:
                return index
            left += num
        return -1