LeetCode747 Largest Number At Least Twice of Others

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博客给出一个整数数组相关题目，需判断数组中最大元素是否至少为其他每个元素的两倍。若是，返回最大元素的索引；否则返回 -1。还给出了两个示例及数组长度和元素值范围等说明。

题目：

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].

public class LargestNumberTwiceOthers {
    public static int dominantIndex(int[] nums) {
        if (nums == null || nums.length == 0)
            return -1;
        int maxValue = -1;
        int maxValueIndex = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > maxValue){
                maxValue = nums[i];
                maxValueIndex = i;
            }
        }

        for (int i = 0; i < nums.length; i++) {
            if (i == maxValueIndex)
                continue;
            else if (maxValue < 2 * nums[i]){
                maxValueIndex = -1;
            }
        }

        return maxValueIndex;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4};
        System.out.println(dominantIndex(nums));
    }
}