747. Largest Number Greater Than Twice of Others

本文介绍了一个简单的算法问题,即在给定整数数组中找到最大的元素是否至少是其他每个元素的两倍。如果是,则返回该元素的索引;否则返回-1。通过遍历数组并维护最大值和次大值来解决此问题。

    这道题为简单题

题目:         

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:

    这个题比较简单,用两个变量跟踪整个列表中的最大和次大值的索引,最后比较即可。

代码:

 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

 

转载于:https://www.cnblogs.com/liuxinzhi/p/8108587.html

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