leetcode-747. Largest Number Greater Than Twice of Others

本文链接：https://blog.youkuaiyun.com/qq_32832023/article/details/78891289

本文介绍了一个简单而实用的算法挑战：如何判断数组中是否存在一个元素，该元素至少是其他所有元素最大值的两倍，并返回其索引。文章提供了一种有效的解决方案，仅需一次遍历即可完成任务。

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747. Largest Number Greater Than 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.

Note:

nums will have a length in the range [1, 50].
Every nums[i] will be an integer in the range [0, 99].

此题做于 leetcode Weekly Contest 64

推荐大家去试一下这个每周比赛~

题意：

给定一个数组，找出数组里是否存在最大的一个数是第二大数的两倍还多。存在返回下标；不存在返回-1

思路：

这题就比较简单了，遍历找到最大的数字保存数字和下标，再遍历找到第二大的数字。按要求返回就可以了。
只需要O(n)

class Solution {
    public int dominantIndex(int[] nums) {
        int max = Integer.MIN_VALUE,flag=0;
        for (int i =0 ;i<nums.length;i++){
            if(nums[i]>=max) {
                max = nums[i];
                flag = i;
            }
        }
        Arrays.sort(nums);
        if(max >= 2*nums[nums.length-2]) return flag;
        return -1;
    }
}