[LeetCode] 747. Largest Number At Least Twice of Others-优快云博客

本文介绍了一种算法，用于在整数数组中找到是否有一个元素至少是其他所有元素值两倍的方法，并返回该元素的索引。如果不存在这样的元素，则返回-1。

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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.

题意：这个数组中一定存在唯一大的一个数，这个数是否比其他的数大两倍
只要判断最大和第二大就行了，因为是纯乱序的，遍历两遍就行了 O(n) 的复杂度

Java

 1 class Solution {
 2     public int dominantIndex(int[] nums) {
 3         int max = 0;
 4         int maxIndex = 0;
 5         int secMax = 0;
 6         for (int i = 0; i < nums.length; i++) {
 7             if (max < nums[i]) {
 8                 max = nums[i];
 9                 maxIndex = i;
10             }
11         }
12         for (int i = 0; i < nums.length; i++) {
13             if (i == maxIndex)
14                 continue;
15             secMax = secMax > nums[i] ? secMax : nums[i];
16         }
17         if ((max - 2*secMax) >= 0)
18             return maxIndex;
19         return -1;
20     }
21 }