题目描述：

LeetCode 747. Largest Number Greater Than Twice of Others

In a given integer array nums, there is always exactly one largest element.

Find whether the largest element in the array is at least twice as much as every other number in the array.

If it is, return the index of the largest element, otherwise return -1.

Example 1:

Input: nums = [3, 6, 1, 0]
Output: 1
Explanation: 6 is the largest integer, and for every other number in the array x,
6 is more than twice as big as x.  The index of value 6 is 1, so we return 1.

Example 2:

Input: nums = [1, 2, 3, 4]
Output: -1
Explanation: 4 isn't at least as big as twice the value of 3, so we return -1.

Note:

1. nums will have a length in the range [1, 50].
2. Every nums[i] will be an integer in the range [0, 99].

题目大意：

给定具有唯一最大值的数组nums，判断其最大值是否不小于其余所有元素的2倍，若成立则返回最大值下标，否则返回-1。

解题思路：

蛮力法（Brute Force）

Python代码：

class Solution(object):
    def dominantIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        maxn = max(nums)
        return [-1, nums.index(maxn)][sum(maxn < 2 * n for n in nums) == 1]

 

本文链接：http://bookshadow.com/weblog/2017/12/24/leetcode-largest-number-greater-than-twice-of-others/
请尊重作者的劳动成果，转载请注明出处！书影博客保留对文章的所有权利。

如果您喜欢这篇博文，欢迎您捐赠书影博客： ，查看支付宝二维码