Python：747.Largest Number At Least Twice of Others代码实例

747.Largest Number At Least 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:

nums will have a length in the range [1, 50].

Every nums[i] will be an integer in the range [0, 99].

思路：

因为输入的都是正数，直接复制一个nums_copy，然后将原列表nums排序，然后判断最大的数是否是第二大数的2倍就可以了，值得注意的是，当列表长度只有1的时候直接输出。

我的代码：

class Solution:
    def dominantIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums_copy=nums[:]
        if len(nums) == 1:
            return 0
        else:
            nums.sort()
            if nums[-1] >= nums[-2] * 2:
                return nums_copy.index(nums[-1])
            else:
                return -1