747_Largest-Number-At-Least-Twice-of-Others

747_Largest-Number-At-Least-Twice-of-Others

目录

• 747_Largest-Number-At-Least-Twice-of-Others
  • Description
  • Solution
    • Java solution
    • Python solution

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:

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

Solution

Java solution

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

Runtime: 17 ms

Python solution

class Solution:
    def dominantIndex(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 1:
            return 0
        
        largest, second_largest, k = 0, 0, 0
        for i, num in enumerate(nums):
            if num > largest:
                second_largest, largest, k = largest, num, i
            elif num > second_largest:
                second_largest = num
        
        if largest >= 2*second_largest:
            return k
        else:
            return -1

Runtime: 44 ms