# 747. Largest Number At Least Twice of Others 至少是其他数字两倍的最大数

@TOC

## # 题目描述

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].

## # 解题方法

### # 寻找两次最大值

``````class Solution(object):
def dominantIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 1:
return 0
largest = max(nums)
ind = nums.index(largest)
nums.pop(ind)
if largest >= 2 * max(nums):
return ind
else:
return -1
``````

### # 排序

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

### # 大顶堆

``````class Solution(object):
def dominantIndex(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 1:
return 0
heap = [(-num, i) for i, num in enumerate(nums)]
heapq.heapify(heap)
largest, ind = heapq.heappop(heap)
if largest <= 2 * heapq.heappop(heap)[0]:
return ind
return -1
``````

## # 日期

2018 年 1 月 28 日 2018 年 11 月 21 日 —— 又是一个美好的开始