leetcode 747. Largest Number At Least Twice of Others

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.

记录，第一大的数和第二大的数，判断2倍关系就行了。

class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        int max1 = -1;
        int max2 = -1;
        int id = 0;
        for(int i = 0; i < nums.size(); ++i) {
            int x = nums[i];
            if (x > max1) max2 = max1, max1 = x, id = i;
            else if (x > max2) max2 = x;
        }
        if (max1 >= 2*max2) return id;
        return -1;
    }
};