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·Apr 26, 2022·

# 747. Largest Number At Least Twice of Others

## 題目敘述

You are given an integer array `nums` where the largest integer is unique.

Determine 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, or return `-1` otherwise.

Example 1:

``````Input: nums = [3,6,1,0]
Output: 1
Explanation: 6 is the largest integer.
For every other number in the array x, 6 is at least 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 is less than twice the value of 3, so we return -1.
``````

Example 3:

``````Input: nums = [1]
Output: 0
Explanation: 1 is trivially at least twice the value as any other number because there are no other numbers.
``````

Constraints:

• `1 <= nums.length <= 50`
• `0 <= nums[i] <= 100`
• The largest element in `nums` is unique.

Hint 1:

Scan through the array to find the unique largest element `m`, keeping track of it's index `maxIndex`. Scan through the array again. If we find some `x != m` with `m < 2*x`, we should return `-1`. Otherwise, we should return `maxIndex`.

## 解法解析

### 解法範例

#### Go

``````func dominantIndex(nums []int) int {
if len(nums) == 0 {
return -1
}

var (
largest       int = -1
secondLargest int = -1
maxIndex      int
)

for i, val := range nums {
if val >= largest {
secondLargest = largest
largest = val
maxIndex = i
} else if val > secondLargest {
secondLargest = val
}
}

if largest >= secondLargest*2 {
return maxIndex
}
return -1
}
``````

#### JavaScript

``````/**
* @param {number[]} nums
* @return {number}
*/
var dominantIndex = function (nums) {
if (nums.length === 0) {
return -1;
}
let largest = -1,
second = -1,
maxIdx = 0;

for (let i = 0; i < nums.length; i++) {
if (nums[i] > largest) {
second = largest;
largest = nums[i];
maxIdx = i;
} else if (nums[i] > second) {
second = nums[i];
}
}

return largest >= second * 2 ? maxIdx : -1;
};
``````

#### Kotlin

``````class Solution {
fun dominantIndex(nums: IntArray): Int {
if (nums.size == 0) return -1

var largest = -1
var secondLargest = -1
var largestIndex = 0

for ((i, value) in nums.withIndex()) {
if (value > largest) {
secondLargest = largest
largest = value
largestIndex = i
} else if (value > secondLargest) {
secondLargest = value
}
}

return if (largest >= secondLargest * 2) largestIndex else -1
}
}
``````

#### PHP

``````class Solution
{

/**
* @param Integer[] \$nums
* @return Integer
*/
function dominantIndex(\$nums)
{
if (count(\$nums) == 0) {
return -1;
}
\$max = max(\$nums);
\$maxIndex = array_search(\$max, \$nums);
foreach (\$nums as \$idx => \$value) {
if (\$idx == \$maxIndex) {
continue;
}
if (\$value * 2 > \$max) {
return -1;
}
}
return \$maxIndex;
}
}
``````

#### Python

``````class Solution:
def dominantIndex(self, nums: List[int]) -> int:
if len(nums) == 0:
return -1

largest = -1
second = -1
maxIdx = 0

for i, val in enumerate(nums):
if val >= largest:
second = largest
largest = val
maxIdx = i
elif val > second:
second = val

return maxIdx if largest >= second * 2 else -1
``````

#### Rust

``````
``````

#### Swift

``````class Solution {
func dominantIndex(_ nums: [Int]) -> Int {
guard nums.count > 1 else { return 0 }

var largest = -1
var secondLargest = -1
var largestIndex = 0

for (i, val) in nums.enumerated() {
if val >= largest {
secondLargest = largest
largest = val
largestIndex = i
} else if val > secondLargest {
secondLargest = val
}
}

return largest >= 2 * secondLargest ? largestIndex : -1
}
}
``````