[leetcode-748-Largest Number At Least Twice of Others]-优快云博客

本文介绍了一种算法，用于在整数数组中找到是否有一个元素至少是其他所有元素值两倍的方法，并给出了具体的实现代码。

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].

思路：

扫描一遍数组，记录最大值与第二大的值，如果最大的是第二大的两倍以上，那么返回索引即可。

 int dominantIndex(vector<int>& nums)
 {
    int len = nums.size();
    int biggest = 0,second = 0,index = 0;
    
    for(int i=0;i<len;i++)
    {
        if(nums[i]>biggest)
        {
            second = biggest;
            biggest = nums[i];
            index = i;
        }
        else if(nums[i]>second)
        {
            second = nums[i];
        }
    }
    if(biggest>=second*2)return index;
    else return -1;        
 }