# 747.Largest Number At Least Twice Of Others

**747.Largest Number At Least Twice Of Others**

难度:Easy

> 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].
```

按题意来即可。

```
class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        int max_ind=max_element(nums.begin(), nums.end()) -nums.begin();
        for(int i=0;i<nums.size();i++)
        {
            if(nums[i]*2<=nums[max_ind]) continue;
            else if(i==max_ind) continue;
            else return -1;
        }
        return  max_ind;
    }
};
```

> 执行用时 :4 ms, 在所有 C++ 提交中击败了88.42%的用户\
> 内存消耗 :8.3 MB, 在所有 C++ 提交中击败了85.06%的用户


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