[Swift]LeetCode453. 最小移動次數使數組元素相等 | Minimum Moves to Equal Array Elements

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Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.

Example:

Input:
[1,2,3]

Output:
3

Explanation:
Only three moves are needed (remember each move increments two elements):

[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]

---

給定一個長度爲 n 的非空整數數組，找到讓數組全部元素相等的最小移動次數。每次移動能夠使 n - 1 個元素增長 1。

示例:

輸入:
[1,2,3]

輸出:
3

解釋:
只須要3次移動（注意每次移動會增長兩個元素的值）：

[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]

---

56ms

 1 class Solution {
 2     func minMoves(_ nums: [Int]) -> Int {
 3         var min = nums.first!
 4         var sum = 0
 5         for num in nums {
 6             sum += num
 7             if num < min {
 8                 min = num
 9             }
10         }
11         return sum - nums.count*min
12     }
13 }

---

84ms

 1 class Solution {
 2     func minMoves(_ nums: [Int]) -> Int {
 3         var minNum = Int.max
 4         var sum = 0
 5         for num in nums {
 6             sum += num
 7             minNum = min(minNum, num)
 8         }
 9 
10         return sum - minNum * nums.count
11     }
12 }

---

88ms

1 class Solution {
2     func minMoves(_ nums: [Int]) -> Int {
3         let min = nums.min()!
4         return nums.reduce(0 ,{$0 + $1}) - min * nums.count
5     }
6 }

---

104ms

 1 class Solution {
 2     func minMoves(_ nums: [Int]) -> Int {
 3         if nums.isEmpty { return 0 }
 4         var mini = nums[0]
 5         
 6         for num in nums {
 7             mini = min(num, mini)
 8         }
 9     
10         var res = 0
11         for num in nums {
12             res += num - mini
13         }
14         return res
15     }
16 }

---

116ms

1 class Solution {
2     func minMoves(_ nums: [Int]) -> Int {
3         let min = nums.min()!
4 
5         return nums.reduce(0) { total, num in total + num - min }
6     }
7 }