747. Largest Number At Least Twice of Others（python+cpp）

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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:
nums will have a length in the range [1, 50].
Every nums[i] will be an integer in the range[0, 99].

解释：
判断数组章最大的数是否是所有其他数字的至少2倍大，如果是，返回气位置，如果不是，返回-1。
排序之后比较最大的数和第二大的数即可。
python代码：

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

c++代码：

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        int n=nums.size();
        if (n<2)
            return 0;
        vector<int>newnums(nums.begin(),nums.end());
        sort(newnums.begin(),newnums.end());
        return newnums[n-1]>=2*newnums[n-2]?(max_element(nums.begin(),nums.end())-nums.begin()):-1;
    }
};

总结：
c++中用max_element()返回数组中最大值的地址。