Largest Number At Least Twice of Others

Posted 2020-10-27 jiadyang

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].

思路还是比较清楚的，比较最大和次大，是否有两倍以上：

 1 class Solution {
 2 public:
 3     int dominantIndex(vector<int>& nums) {
 4         
 5         auto i = max_element(nums.begin(), nums.end());
 6         
 7         sort(nums.begin(), nums.end());
 8         
 9         if(*(nums.end() - 2) == 0 && *(nums.end() - 1) != 0)
10             return i - nums.begin();
11         
12         if( *(nums.end() - 1) / *(nums.end() - 2) >= 2 ? true : false)
13         {
14             return i - nums.begin();
15         }
16         return -1;
17     }
18 };

再贴一个没用到sort的解，摘自discuss：

 1 class Solution {
 2 public:
 3     int dominantIndex(vector<int>& nums) {
 4         int maxn = INT_MIN, idx = -1, sec = INT_MIN;
 5         for (int i = 0; i < nums.size(); i++) {
 6             if (nums[i] > maxn) {
 7                 sec = maxn;
 8                 maxn = nums[i];
 9                 idx = i;
10             } else if(nums[i] > sec){
11                 sec = nums[i];
12             }
13         }
14         return sec * 2 > maxn ? -1 : idx;
15     }
16 };