748. Largest Number At Least Twice of Others

Posted 2020-10-19 __Meng

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:

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

 

查找数组中最大的元素是否至少是数组中其他数字的两倍。是的话就返回最大元素的索引。

 

C++（6ms）：

 1 class Solution {
 2 public:
 3     int dominantIndex(vector<int>& nums) {
 4         int len = nums.size() ;
 5         if (len < 2)
 6             return 0 ;
 7         int maxOne = 0 ;
 8         int maxTwo = 0 ;
 9         int index = 0 ;
10         for(int i = 0 ; i < len ; i++){
11             if (nums[i] > maxOne){
12                 maxTwo = maxOne ;
13                 maxOne = nums[i] ;
14                 index = i ;
15             }else if(nums[i] > maxTwo){
16                 maxTwo = nums[i] ;
17             }
18         }
19         return maxTwo*2 <= maxOne ? index : -1 ;
20     }
21 };