747. Largest Number At Least Twice of Others

问题：

给定数组，若存在最大值是任意其他元素的2倍以上，则返回该最大值的index，否则返回-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].

　　

解法：

设两个变量记录最大值largest和第二大值second

遍历数组，最后若第二大值*2<=最大值，那么返回largest的index

代码参考：

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        int res=-1, largest=0, second=0;
        for(int i=0; i<nums.size(); i++){
            if(nums[i]>largest){
                second=largest;
                largest=nums[i];
                res=i;
            }else if(nums[i]>second && nums[i] != largest){
                second=nums[i];
            }
        }
        return second*2<=largest? res:-1;
    }
};