[Swift]LeetCode453. 最小移动次数使数组元素相等 | 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]

---

给定一个长度为 n 的非空整数数组，找到让数组所有元素相等的最小移动次数。每次移动可以使 n - 1 个元素增加 1。

示例:

输入:
[1,2,3]

输出:
3

解释:
只需要3次移动（注意每次移动会增加两个元素的值）：

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

---

56ms

class Solution {
    func minMoves(_ nums: [Int]) -> Int {
        var min = nums.first!
        var sum = 0
        for num in nums {
            sum += num
            if num < min {
                min = num
            }
        }
        return sum - nums.count*min
    }
}

---

84ms

class Solution {
    func minMoves(_ nums: [Int]) -> Int {
        var minNum = Int.max
        var sum = 0
        for num in nums {
            sum += num
            minNum = min(minNum, num)
        }

        return sum - minNum * nums.count
    }
}

---

88ms

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

---

104ms

class Solution {
    func minMoves(_ nums: [Int]) -> Int {
        if nums.isEmpty { return 0 }
        var mini = nums[0]
        
        for num in nums {
            mini = min(num, mini)
        }
    
        var res = 0
        for num in nums {
            res += num - mini
        }
        return res
    }
}

---

116ms

class Solution {
    func minMoves(_ nums: [Int]) -> Int {
        let min = nums.min()!

        return nums.reduce(0) { total, num in total + num - min }
    }
}