747. Largest Number At Least Twice of Others

문제

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.

정수의 배열이 주어질 때, 가장 큰 수가 다른 수의 2배 이상이 되는지 확인하라.

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].

---

Solution

1. 먼저 가장 큰 수의 인덱스를 maxIndex에 저장한다.
2. 그 인덱스를 제외한 다른 요소들에 2를 곱해 max값보다 크면 -1을 반환한다.
3. 아니면 maxIndex를 반환한다.

class Solution {
    public int dominantIndex(int[] nums) {
        int maxIndex=0;
        for (int i = 1; i < nums.length; i++){
            if (nums[i] > nums[maxIndex]){
                maxIndex = i;
            }
        }
        
        for (int i = 0; i < nums.length; i++){
            if (maxIndex != i && nums[i]*2 > nums[maxIndex]){
                return -1;
            }
        }
        
        return maxIndex;
    }
}