747.Largest Number At Least Twice of Others

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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].

class Solution:
    def dominantIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums)<1:
            return -1
        if len(nums)==1:
            return 0
        first,second,pos = -1000000,-1000000,-1
        for i in range(len(nums)):
            if nums[i]>first:
                second = first
                first = nums[i]
                pos = i
                continue
            if nums[i]>second:
                second = nums[i]
        if first>=second*2:
            return pos
        return -1

747.Largest Number At Least Twice of Others

标签：continue   nan   range   for   turn   always   note   ISE   sel   

原文地址：https://www.cnblogs.com/bernieloveslife/p/9771804.html