LeetCode-Largest Number At Least Twice of Others

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Description:
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].

题意：给定一个一维数组，如果数组中的最大的数是所有其他元素的两倍及以上，返回其下标，没有，则返回-1；

解法一：首先第一步我们要找到那个最大的数，之后遍历一遍数组，比较其是否满足是所有其他元素的两倍及以上；

Java

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

解法二：我们可以找到数组中最大的数numFirst，和数组中次大的数numSecond，如果满足numFirst >= numSecond * 2，那么对于numFirst来说他就会比数组中所有的其他元素都大两倍及以上；

Java

class Solution {
    public int dominantIndex(int[] nums) {
        int maxFirst = 0;
        int maxSecond = 0;
        int maxNumIndex = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > maxFirst) {
                maxSecond = maxFirst;
                maxFirst = nums[i];
                maxNumIndex = i;
            } else if (nums[i] > maxSecond) {
                maxSecond = nums[i];
            }
        }
        return maxFirst >= maxSecond * 2 ? maxNumIndex : -1;
    }
}