[Swift]LeetCode747. 至少是其餘數字兩倍的最大數 | Largest Number At Least Twice of Others

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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. 

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].

---

在一個給定的數組nums中，老是存在一個最大元素 。

查找數組中的最大元素是否至少是數組中每一個其餘數字的兩倍。

若是是，則返回最大元素的索引，不然返回-1。

示例 1:

輸入: nums = [3, 6, 1, 0]
輸出: 1
解釋: 6是最大的整數, 對於數組中的其餘整數,
6大於數組中其餘元素的兩倍。6的索引是1, 因此咱們返回1. 

示例 2:

輸入: nums = [1, 2, 3, 4]
輸出: -1
解釋: 4沒有超過3的兩倍大, 因此咱們返回 -1. 

提示:

1. nums 的長度範圍在[1, 50].
2. 每一個 nums[i] 的整數範圍在 [0, 99].

---

Runtime: 12 ms

Memory Usage: 18.7 MB

 1 class Solution {
 2     func dominantIndex(_ nums: [Int]) -> Int {
 3         var mx:Int = Int.min
 4         var mxId:Int = 0
 5         for i in 0..<nums.count
 6         {
 7             if mx < nums[i]
 8             {
 9                 mx = nums[i]
10                 mxId = i
11             }
12         }
13         for num in nums
14         {
15             if mx != num && mx - num < num
16             {
17                 return -1
18             }
19         }
20         return mxId
21     }
22 }

---

12ms

1 class Solution {
2     func dominantIndex(_ nums: [Int]) -> Int {
3     let maximumNumber = nums.max()!
4     let index = nums.firstIndex(of: maximumNumber)!
5      let _nums = nums.filter{$0 * 2 > maximumNumber}
6        return _nums.count > 1 ? -1 : index        
7     }
8 }

---

16ms

 1 class Solution {
 2     func dominantIndex(_ nums: [Int]) -> Int {
 3         guard nums.count > 1 else {
 4             return 0
 5         }
 6         
 7        let ar = nums.enumerated().sorted{$0.1 > $1.1}
 8         if ar[0].1 >= 2*ar[1].1 {
 9             return ar[0].0
10         }
11         return -1
12     }
13 }

---

16ms

 1 class Solution {
 2     func dominantIndex(_ nums: [Int]) -> Int {
 3         if nums.count == 1 {
 4             return 0
 5         }
 6         
 7         var index: Int = 0
 8         var maxValue: Int = 0
 9         var secondValue: Int = 0
10     
11         for i in 0...nums.count - 1{            
12             if nums[i] > maxValue{
13                 secondValue = maxValue
14                 maxValue = nums[i]
15                 index = i
16             }else if nums[i] > secondValue{
17                 secondValue = nums[i]
18             }           
19         }
20         
21         return maxValue >= (secondValue * 2) ? index : -1
22     }
23 }

---

20ms

 1 class Solution {
 2     
 3     var greatestNumber = (index: -1, value: -1)
 4     var secondGreatestNumber = (index: -1, value: -1)
 5     var currentNumber = (index: -1, value: -1)
 6     
 7     func dominantIndex(_ nums: [Int]) -> Int {
 8 
 9         findTwoGreatestNumbers(in: nums)
10         if greatestNumber.value >= secondGreatestNumber.value * 2 {
11             return greatestNumber.index
12         } else {
13             return -1
14         }
15     }
16     
17     private func findTwoGreatestNumbers(in nums: [Int]) {
18         for (index, value) in nums.enumerated() {
19             currentNumber = (index,value)
20             if currentNumber.value > greatestNumber.value {
21                 secondGreatestNumber = greatestNumber
22                 greatestNumber = currentNumber
23             } else if currentNumber.value > secondGreatestNumber.value {
24                 secondGreatestNumber = currentNumber
25             }
26         }
27     }
28 }

---

24ms

 1 class Solution {
 2     func dominantIndex(_ nums: [Int]) -> Int {
 3         guard !nums.isEmpty else {
 4             return -1
 5         }
 6         
 7         var maxIndex = 0
 8         for index in 1..<nums.count {
 9             let num = nums[index]
10             if num > nums[maxIndex] {
11                 maxIndex = index
12             }
13         }
14         
15         let maxNum = nums[maxIndex]
16         for num in nums where num != 0 && num != maxNum {
17             if maxNum / num < 2 {
18                 return -1
19             }
20         }
21         
22         return maxIndex
23     }
24 }

---

28ms

1 class Solution {
2     func dominantIndex(_ nums: [Int]) -> Int {
3     guard let largestNumber = nums.sorted(by: >).first else { return -1 }
4     let dominateNums = nums.filter({ $0 != largestNumber && largestNumber < ($0 * 2) })
5     return dominateNums.isEmpty ? nums.index(of: largestNumber)! : -1
6    }
7 }