這道題為簡單題

題目:　　　　　　　　　

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].

思路：

　　　　這個題比較簡單，用兩個變量跟蹤整個列表中的最大和次大值的索引，最後比較即可。

代碼：

 1 class Solution(object):
 2     def dominantIndex(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 

 7         
 8         
 9         max_index = 0
10         min_index = float(‘-inf‘)
11         if len(nums) < 2:
12             return max_index
13         
14         for i in range(1, len(nums)):
15             if nums[i] > nums[max_index]:
16                 min_index = max_index
17                 max_index = i
18             elif (nums[i] < nums[max_index]) and (min_index == float(‘-inf‘) or nums[i] > nums[min_index]):
19                 min_index = i;
20         
21         if nums[max_index] >= nums[min_index] * 2: return max_index
22         else: return -1

747. Largest Number Greater Than Twice of Others